Notational confusion of multivariable derivatives - Revision history

https://calculus.subwiki.org/w/index.php?action=history&feed=atom&title=Notational_confusion_of_multivariable_derivatives Notational confusion of multivariable derivatives - Revision history 2026-10-18T16:39:08Z Revision history for this page on the wiki MediaWiki 1.41.2 https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3211&oldid=prev IssaRice at 23:43, 30 May 2020 2020-05-30T23:43:41Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 23:43, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l2">Line 2:</td> <td colspan="2" class="diff-lineno">Line 2:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* The thing where <math>\frac{\partial w}{\partial t}</math> can mean two different things on LHS and RHS when <math>t</math> is used as both an initial and intermediate variable. (See Folland for details.)</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* The thing where <math>\frac{\partial w}{\partial t}</math> can mean two different things on LHS and RHS when <math>t</math> is used as both an initial and intermediate variable. (See Folland for details.)</div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">* the meaning of partial derivatives depends on which variables you take to be independent; see p. 75 of folland</ins></div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* The thing where if <math>f(x,y) = (x^2,y^2)</math> then <math>\frac{\partial f}{\partial x}(x,x)</math> feels like it might be <math>(2x,2x)</math> even though it's actually <math>(2x,0)</math>. (Example from Tao.) See also [https://issarice.com/mathematics-and-notation]</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* The thing where if <math>f(x,y) = (x^2,y^2)</math> then <math>\frac{\partial f}{\partial x}(x,x)</math> feels like it might be <math>(2x,2x)</math> even though it's actually <math>(2x,0)</math>. (Example from Tao.) See also [https://issarice.com/mathematics-and-notation]</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* The ambiguity of expressions like <math>\nabla f(Ax)</math></div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* The ambiguity of expressions like <math>\nabla f(Ax)</math></div></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3210&oldid=prev IssaRice at 23:07, 30 May 2020 2020-05-30T23:07:30Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 23:07, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8">Line 8:</td> <td colspan="2" class="diff-lineno">Line 8:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Working off the example from Tao above, let <math>f(x,y) = (x^2,y^2)</math>. What does <math>\frac{d}{dx} f(x,x)</math> mean? Here are four possibilities:</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Working off the example from Tao above, let <math>f(x,y) = (x^2,y^2)</math>. What does <math>\frac{d}{dx} f(x,x)</math> mean? Here are four possibilities:</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) <del style="font-weight: bold; text-decoration: none;">evaluated </del>at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Since our point is <math>(x_0, y_0) = (x,x)</math>, we have <math>\begin{pmatrix}2x & 0 \\ 0 & 2x\end{pmatrix}</math>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Since our point is <math>(x_0, y_0) = (x,x)</math>, we have <math>\begin{pmatrix}2x & 0 \\ 0 & 2x\end{pmatrix}</math>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3209&oldid=prev IssaRice at 23:07, 30 May 2020 2020-05-30T23:07:09Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 23:07, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8">Line 8:</td> <td colspan="2" class="diff-lineno">Line 8:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Working off the example from Tao above, let <math>f(x,y) = (x^2,y^2)</math>. What does <math>\frac{d}{dx} f(x,x)</math> mean? Here are four possibilities:</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Working off the example from Tao above, let <math>f(x,y) = (x^2,y^2)</math>. What does <math>\frac{d}{dx} f(x,x)</math> mean? Here are four possibilities:</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. <del style="font-weight: bold; text-decoration: none;">Applying this linear map to </del>(x,x), we <del style="font-weight: bold; text-decoration: none;">get </del><math>\begin{pmatrix}<del style="font-weight: bold; text-decoration: none;">2x_0 </del>& 0 \\ 0 & <del style="font-weight: bold; text-decoration: none;">2y_0</del>\end{pmatrix}<del style="font-weight: bold; text-decoration: none;">\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</del></math>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. <ins style="font-weight: bold; text-decoration: none;">Since our point is <math>(x_0, y_0) = </ins>(x,x)<ins style="font-weight: bold; text-decoration: none;"></math></ins>, we <ins style="font-weight: bold; text-decoration: none;">have </ins><math>\begin{pmatrix}<ins style="font-weight: bold; text-decoration: none;">2x </ins>& 0 \\ 0 & <ins style="font-weight: bold; text-decoration: none;">2x</ins>\end{pmatrix}</math>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3208&oldid=prev IssaRice at 22:50, 30 May 2020 2020-05-30T22:50:33Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 22:50, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l6">Line 6:</td> <td colspan="2" class="diff-lineno">Line 6:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* dual basis stuff -- see Tao's explanation of this in p. 225 of [https://terrytao.files.wordpress.com/2011/06/blog-book.pdf]</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* dual basis stuff -- see Tao's explanation of this in p. 225 of [https://terrytao.files.wordpress.com/2011/06/blog-book.pdf]</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Working off the example from Tao above, let <math>f(x,y) = (x^2,y^2)</math>. What does <math>\frac{d}{dx} f(x,x)</math> mean? Here are <del style="font-weight: bold; text-decoration: none;">three </del>possibilities:</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Working off the example from Tao above, let <math>f(x,y) = (x^2,y^2)</math>. What does <math>\frac{d}{dx} f(x,x)</math> mean? Here are <ins style="font-weight: bold; text-decoration: none;">four </ins>possibilities:</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3207&oldid=prev IssaRice at 22:49, 30 May 2020 2020-05-30T22:49:54Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 22:49, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l11">Line 11:</td> <td colspan="2" class="diff-lineno">Line 11:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># There's implicitly a function <math>\phi(x,y) = (x,x)</math>, so <math>\frac{d}{dx} f(x,x) = \frac{d}{dx} f(\phi(x,y))</math>. Using the chain rule, this is <math>(f\circ \phi)'(x,y) = f'(\phi(x,y))\phi'(x,y) = \left.\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\right\vert_{(x_0,y_0) = \phi(x,y)} \begin{pmatrix}1 & 0 \\ 1 & 0\end{pmatrix}</math></div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># There's implicitly a function <math>\phi(x,y) = (x,x)</math>, so <math>\frac{d}{dx} f(x,x) = \frac{d}{dx} f(\phi(x,y))</math>. Using the chain rule, this is <math>(f\circ \phi)'(x,y) = f'(\phi(x,y))\phi'(x,y) = \left.\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\right\vert_{(x_0,y_0) = \phi(x,y)} \begin{pmatrix}1 & 0 \\ 1 <ins style="font-weight: bold; text-decoration: none;">& 0\end{pmatrix} = \begin{pmatrix}2x & 0 \\ 2x </ins>& 0\end{pmatrix}</math><ins style="font-weight: bold; text-decoration: none;">.</ins></div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Big picture==</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Big picture==</div></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3206&oldid=prev IssaRice at 22:48, 30 May 2020 2020-05-30T22:48:28Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 22:48, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l11">Line 11:</td> <td colspan="2" class="diff-lineno">Line 11:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to (2x,2x).</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> <tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"># There's implicitly a function <math>\phi(x,y) = (x,x)</math>, so <math>\frac{d}{dx} f(x,x) = \frac{d}{dx} f(\phi(x,y))</math>. Using the chain rule, this is <math>(f\circ \phi)'(x,y) = f'(\phi(x,y))\phi'(x,y) = \left.\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\right\vert_{(x_0,y_0) = \phi(x,y)} \begin{pmatrix}1 & 0 \\ 1 & 0\end{pmatrix}</math></ins></div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Big picture==</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Big picture==</div></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3204&oldid=prev IssaRice at 09:25, 30 May 2020 2020-05-30T09:25:27Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:25, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9">Line 9:</td> <td colspan="2" class="diff-lineno">Line 9:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to <del style="font-weight: bold; text-decoration: none;">the following</del>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related. If we assume the relationship y=x then this reduces to <ins style="font-weight: bold; text-decoration: none;">(2x,2x)</ins>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3203&oldid=prev IssaRice at 09:25, 30 May 2020 2020-05-30T09:25:11Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:25, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9">Line 9:</td> <td colspan="2" class="diff-lineno">Line 9:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>. We can't evaluate this further since we don't know how x and y are related<ins style="font-weight: bold; text-decoration: none;">. If we assume the relationship y=x then this reduces to the following</ins>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3200&oldid=prev IssaRice at 09:12, 30 May 2020 2020-05-30T09:12:50Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:12, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9">Line 9:</td> <td colspan="2" class="diff-lineno">Line 9:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math>.</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x). We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math><ins style="font-weight: bold; text-decoration: none;">. We can't evaluate this further since we don't know how x and y are related</ins>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> </table>

IssaRice https://calculus.subwiki.org/w/index.php?title=Notational_confusion_of_multivariable_derivatives&diff=3199&oldid=prev IssaRice at 09:07, 30 May 2020 2020-05-30T09:07:33Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:07, 30 May 2020</td> </tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9">Line 9:</td> <td colspan="2" class="diff-lineno">Line 9:</td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># It's <math>f'</math> (i.e., the total derivative of f) evaluated at the point (x,x). Once we fix a point <math>(x_0,y_0)</math>, then <math>f'(x_0,y_0)</math> is a linear map <math>\mathbf R^2 \to \mathbf R^2</math> defined by the matrix <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}</math>. Applying this linear map to (x,x), we get <math>\begin{pmatrix}2x_0 & 0 \\ 0 & 2y_0\end{pmatrix}\begin{pmatrix}x \\ x\end{pmatrix} = \begin{pmatrix}2x_0x \\ 2y_0x\end{pmatrix} \in \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x).</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div># It's <math>\frac{df}{dx}</math> (i.e., the total derivative of f with respect to x, in which we treat the variable y as a function of x, and use the chain rule <math>\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}</math> to compute) evaluated at the point (x,x)<ins style="font-weight: bold; text-decoration: none;">. We have <math>(2x,0) + (0,2y)\frac{dy}{dx} = (2x, 2y\frac{dy}{dx})</math></ins>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div># We first compute f(x,x) to get an expression involving only x, which implicitly defines a function <math>\mathbf R \to \mathbf R^2</math>. We now differentiate this function. The result is the function <math>x \mapsto (2x,2x) : \mathbf R \to \mathbf R^2</math>.</div></td></tr> <tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr> </table>

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