Quiz:Chain rule for differentiation: Difference between revisions
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- <math>(f_1 \cdot f_2' \cdot f_3') \circ (f_2 \cdot f_3') \circ f_3</math> | - <math>(f_1 \cdot f_2' \cdot f_3') \circ (f_2 \cdot f_3') \circ f_3</math> | ||
- <math>f_1' \circ f_2' \circ f_3'</math> | - <math>f_1' \circ f_2' \circ f_3'</math> | ||
{ Suppose <math>f</math> is a differentiable function from <math>\R</math> to <math>\R</math> and <math>a, b \in \R</math> are such that <math>f(a) = a</math> and <math>f'(a) = b</math>. What is the value of <math>(f \circ f \circ \dots \circ f)'(a)</math>, where <math>\circ</math> denotes the [[composite of two functions]] and <math>f</math> occurs <math>n</math> times in the expression, with <math>n \ge 3</math>? | |||
|type="()"} | |||
- <math>a^n</math> | |||
- <math>a^{n - 1}b</math> | |||
- <math>a^{n - 1}b + ab^{n - 1}</math> | |||
- <math>ab^{n - 1}</math> | |||
+ <math>b^n</math> | |||
|| The chain rule gives the derivative as a product of <math>n</math> terms, each of which is of the form <math>f'<math> applied to <math>k</math> iterates of <math>f</math>, with <math>k</math> varying from <math>0</math> to <math>n - 1</math>. Evaluating at <math>a</math> and using <math>f'(a) = a</math>, each term simplifies to <math>f'(a)</math> and hence to <math>b</math>. As there are <math>n</math> such terms, the product is <math>b^n</math>. Note that <math>n \ge 3</math> is not necessary (this reasoning works for <math>n = 1</math> and <math>n = 2</math> as well). That condition was added primarily to dissuade people from using <math>n = 1</math> or <math>n = 2</math> to figure out the correct answer by a process of elimination. | |||
</quiz> | </quiz> |
Revision as of 05:37, 5 December 2023
See chain rule for differentiation and chain rule for higher derivatives for background information.
See Quiz:Differentiation rules for a quiz on all the differentiation rules together.
Formulas