Chain rule for higher derivatives: Difference between revisions

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{{differentiation rule}}
==Statement==
==Statement==


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! Value of <math>n</math> !! Formula for <math>n^{th}</math> derivative of <math>f \circ g</math> at <math>x_0</math>
! Value of <math>n</math> !! Formula for <math>n^{th}</math> derivative of <math>f \circ g</math> at <math>x_0</math>
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| 1 || <math>f'(g(x_0))g'(x_0)</math> (this is the [[chain rule for differentiation]])
| 0 || <math>\! f(g(x_0)) </math> (taking the 0th derivative means doing nothing)
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| 1 || <math>\! f'(g(x_0))g'(x_0)</math> (this is the [[chain rule for differentiation]])
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| 2 || <math>f''(g(x_0))(g'(x_0))^2 + f'(g(x_0))g''(x_0)</math> (obtained by using the [[chain rule for differentiation]] twice ''and'' using the [[product rule for differentiation]]).
| 2 || <math>\! f''(g(x_0))(g'(x_0))^2 + f'(g(x_0))g''(x_0)</math> (obtained by using the [[chain rule for differentiation]] twice ''and'' using the [[product rule for differentiation]])
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Latest revision as of 02:21, 5 December 2023

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

Suppose n is a natural number, and f and g are functions such that g is n times differentiable at x=x0 and f is n times differentiable at g(x0). Then, fg is n times differentiable at x0. Further, the value of the nth derivative is given by a complicated formula involving compositions, products, derivatives, evaluations, and sums that depends on n.

Particular cases

Value of n Formula for nth derivative of fg at x0
0 f(g(x0)) (taking the 0th derivative means doing nothing)
1 f(g(x0))g(x0) (this is the chain rule for differentiation)
2 f(g(x0))(g(x0))2+f(g(x0))g(x0) (obtained by using the chain rule for differentiation twice and using the product rule for differentiation)