Chain rule for differentiation: Difference between revisions

Revision as of 23:21, 15 October 2011

This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
View other differentiation rules

Statement for two functions

The chain rule is stated in many versions:

Version type	Statement
specific point, named functions	Suppose $f$ and $g$ are functions such that $g$ is differentiable at a point $x = x_{0}$ , and $f$ is differentiable at $g (x_{0})$ . Then the composite $f \circ g$ is differentiable at $x_{0}$ , and we have: $\frac{d}{d x} [f (g (x))] \|_{x = x_{0}} = f^{'} (g (x_{0})) g^{'} (x_{0})$
generic point, named functions, point notation	Suppose $f$ and $g$ are functions of one variable. Then, we have $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) g^{'} (x)$ wherever the right side expression makes sense.
generic point, named functions, point-free notation	Suppose $f$ and $g$ are functions of one variable. Then, $(f \circ g)^{'} = (f^{'} \circ g) \cdot g^{'}$ where the right side expression makes sense, where $\cdot$ denotes the pointwise product of functions.
pure Leibniz notation	Suppose $u = g (x)$ is a function of $x$ and $v = f (u)$ is a function of $u$ . Then, $\frac{d v}{d x} = \frac{d v}{d u} \frac{d u}{d x}$

Related rules

Chain rule for higher derivatives
Product rule for differentiation
Product rule for higher derivatives
Differentiation is linear
Inverse function theorem (gives formula for derivative of inverse function).

@@ Line 3: / Line 3: @@
 ==Statement for two functions==
-Suppose <math>f</math> and <math>g</math> are [[function]]s such that <math>g</math> is differentiable at a point <math>x = x_0</math>, and <math>f</math> is differentiable at <math>g(x_0)</math>. Then the [[fact about::composite of two functions|composite]] <math>f \circ g</math> is differentiable at <math>x_0</math>, and we have:
+The chain rule is stated in many versions:
-<math>\frac{d}{dx}[f(g(x))]|_{x = x_0} = f'(g(x_0))g'(x_0)</math>
-In terms of general expressions:
-<math>\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)</math>
-In point-free notation, we have:
-<math>(f \circ g)' = (f' \circ g) \cdot g'</math>
-where <math>\cdot</math> denotes the [[pointwise product of functions]].
+{| class="sortable" border="1"
+! Version type !! Statement
+|-
+| specific point, named functions || Suppose <math>f</math> and <math>g</math> are [[function]]s such that <math>g</math> is differentiable at a point <math>x = x_0</math>, and <math>f</math> is differentiable at <math>g(x_0)</math>. Then the [[fact about::composite of two functions|composite]] <math>f \circ g</math> is differentiable at <math>x_0</math>, and we have: <br><math>\! \frac{d}{dx}[f(g(x))]|_{x = x_0} = f'(g(x_0))g'(x_0)</math>
+|-
+| generic point, named functions, point notation || Suppose <math>f</math> and <math>g</math> are [[function]]s of one variable. Then, we have <br><math>\! \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)</math> wherever the right side expression makes sense.
+|-
+| generic point, named functions, point-free notation || Suppose <math>f</math> and <math>g</math> are [[function]]s of one variable. Then,<br><math>\! (f \circ g)' = (f' \circ g) \cdot g'</math> where the right side expression makes sense, where <math>\cdot</math> denotes the [[pointwise product of functions]].
+|-
+| pure Leibniz notation || Suppose <math>u = g(x)</math> is a function of <math>x</math> and <math>v = f(u)</math> is a function of <math>u</math>. Then, <br><math> \frac{dv}{dx} = \frac{dv}{du}\frac{du}{dx}</math>
+|}
 ==Related rules==