Chain rule for higher derivatives: Difference between revisions

Revision as of 13:42, 26 August 2011

Statement

Suppose $n$ is a natural number, and $f$ and $g$ are functions such that $g$ is $n$ times differentiable at $x=x_{0}$ and $f$ is $n$ times differentiable at $g(x_{0})$ . Then, $f\circ g$ is $n$ times differentiable at $x_{0}$ . Further, the value of the $n^{th}$ 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 $n^{th}$ derivative of $f\circ g$ at $x_{0}$
1	$\!f'(g(x_{0}))g'(x_{0})$ (this is the chain rule for differentiation)
2	$\!f''(g(x_{0}))(g'(x_{0}))^{2}+f'(g(x_{0}))g''(x_{0})$ (obtained by using the chain rule for differentiation twice and using the product rule for differentiation).

@@ Line 8: / Line 8: @@
 ! Value of <math>n</math> !! Formula for <math>n^{th}</math> derivative of <math>f \circ g</math> at <math>x_0</math>
 |-
-| 1 || <math>f'(g(x_0))g'(x_0)</math> (this is the [[chain rule for differentiation]])
+| 1 || <math>\! f'(g(x_0))g'(x_0)</math> (this is the [[chain 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]]).
+| 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]]).
 |}