Quiz:Product rule for differentiation: Difference between revisions
No edit summary |
No edit summary |
||
Line 47: | Line 47: | ||
==Computational feasibility== | ==Computational feasibility== | ||
{{ | <quiz display=simple> | ||
{Suppose <math>f</math> and <math>g</math> are both defined and differentiable at the point 1. Suppose <math>\! f(1) = 2, g(1) = 5, f'(1) = 4, g'(1) = 11</math>. What is the value of <math>(f \cdot g)'(1)</math> where <math>f \cdot g</math> denotes the [[pointwise product of functions]]? | |||
|type="()"} | |||
+ 42 | |||
- 44 | |||
- 54 | |||
- 63 | |||
- The information given is insufficient to find <math>(f \cdot g)'(1)</math>. | |||
{What is the derivaitve of the function <math>x \mapsto \exp(x) \sin x</math>? Hint for derivatives of individual functions: <toggledisplay>Derivative of <math>\exp</math> is <math>\exp</math>, and derivative of <math>\sin</math> is <math>\cos</math>.</toggledisplay> | |||
|type="()"} | |||
+ <math>x \mapsto \exp(x)(\sin x + \cos x)</math> | |||
- <math>x \mapsto \exp(x)(\cos x - \sin x)</math> | |||
- <math>x \mapsto \exp(x)(\sin x - \cos x)</math> | |||
- <math>x \mapsto \exp(x)\cos x + \exp(1) \sin x</math> | |||
- <math>x \mapsto \exp(x)\sin x + \exp(1) \cos x</math> | |||
{What is the derivaitve of the function <math>x \mapsto \sqrt{x}\sin(x^2)</math> for <math>x > 0</math>? This question also requires use of [[chain rule for differentiation]]. | |||
|type="()"} | |||
- <math>x \mapsto \cos(x^2)/(2\sqrt{x})</math> | |||
- <math>x \mapsto 2\sqrt{x}\cos(x^2)</math> | |||
- <math>x \mapsto 2\sqrt{x}(\cos(x^2 + \sin(x^2))</math> | |||
- <math>x \mapsto 2x\sqrt{x}\sin(x^2) + \cos(x^2)/(2 \sqrt{x})</math> | |||
+ <math>x \mapsto 2x\sqrt{x}\cos(x^2) + \sin(x^2)/(2 \sqrt{x})</math> | |||
</quiz> | |||
==Computational results== | |||
<quiz display=simple> | |||
{Suppose <math>f</math> and <math>g</math> are infinitely differentiable functions on all of <math>\R</math> such that both <math>f'</math> and <math>g'</math> are [[periodic function]]s with the same period <math>h > 0</math>. What can we conclude about <math>f \cdot g</math>? | |||
|type="()"} | |||
- <math>f \cdot g</math> must be periodic | |||
- <math>(f \cdot g)'</math> must be periodic, but <math>f \cdot g</math> may or may not be periodic. | |||
- <math>(f \cdot g)''</math> must be periodic, but <math>(f \cdot g)'</math> may or may not be periodic. | |||
- <math>(f \cdot g)'''</math> must be periodic, but <math>(f \cdot g)''</math> may or may not be periodic. | |||
+ We cannot conclude from the given information whether any of the derivatives of <math>f \cdot g</math> is periodic. | |||
{Suppose <math>f</math> and <math>g</math> are functions defined and differentiable on the open interval <math>(0,1)</math>. Suppose, further, that on <math>(0,1)</math>, the derivative functions <math>\! f'</math> and <math>\! g'</math> are both expressible as [[rational function]]s. What can we say about <math>f \cdot g</math> and <math>(f \cdot g)'</math> on <math>(0,1)</math>? | |||
|type="()"} | |||
- Both <math>f \cdot g</math> and <math>(f \cdot g)'</math> are expressible as rational functions. | |||
- <math>f \cdot g</math> is expressible as a rational function, but <math>(f \cdot g)'</math> need not be expressible as a rational function. | |||
- <math>(f \cdot g)'</math> is expressible as a rational function, but <math>f \cdot g</math> need not be expressible as a rational function. | |||
+ Neither <math>f \cdot g</math> nor <math>(f \cdot g)'</math> need be expressible as a rational function. | |||
</quiz> |
Revision as of 21:11, 5 December 2011
For a quiz that tests all the differentiation rules together, see Quiz:Differentiation rules.
For background, see product rule for differentiation and product rule for higher derivatives.
Formulas
Qualitative and existential significance
Computational feasibility
Computational results