proof of chain rule

Let’s say that $g$ is differentiable in $x_0$ and $f$ is differentiable in $y_0=g(x_0)$. We define:

Since $f$ is differentiable in $y_0$, $\phi$ is continuous. We observe that, for $x\ne x_0$,

$$\frac{f(g(x))-f(g(x_0))}{x-x_0}=\phi (g(x))\frac{g(x)-g(x_0)}{x-x_0},$$

in fact, if $g(x)\ne g(x_0)$, it follows at once from the definition of $\phi$, while if $g(x)=g(x_0)$, both members of the equation are 0.

Since $g$ is continuous in $x_0$, and $\phi$ is continuous in $y_0$,

$$\underset{x\to x_0}{lim}\phi (g(x))=\phi (g(x_0))=f^{\prime }(g(x_0)),$$

hence

${(f\circ g)}^{\prime }(x_0)$ $=$ $\underset{x\to x_0}{lim}{\displaystyle \frac{f(g(x))-f(g(x_0))}{x-x_0}}$

$=$ $\underset{x\to x_0}{lim}\phi (g(x)){\displaystyle \frac{g(x)-g(x_0)}{x-x_0}}$

$=$ $f^{\prime }(g(x_0))g^{\prime }(x_0).$

Title	proof of chain rule
Canonical name	ProofOfChainRule
Date of creation	2013-03-22 12:41:48
Last modified on	2013-03-22 12:41:48
Owner	n3o (216)
Last modified by	n3o (216)
Numerical id	6
Author	n3o (216)
Entry type	Proof
Classification	msc 26A06