How I do I prove the Chain Rule for derivatives?

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Answer 1 · 2014-09-21T15:19:52

Before the proof, let us first make the following observation.

By Mean Value Theorem, there exists $c$ in $(x,x+h)$ such that

${g(x+h)-g(x)}/h=g'(c) Leftrightarrow g(x+h)=g(x)+g'(c)h$

Let $t=g'(c)h$ , we have $g(x+h)=g(x)+t$

Note that as $h to 0$ , $t to 0$ .

Proof of Chain Rule

$[f(g(x))]'=lim_{h to infty}{f(g(x+h))-f(g(x))}/{h}$

by multiplying the numerator and the denominator by $g(x+h)-g(x)$ ,

$=lim_{h to 0}{f(g(x+h))-f(g(x))}/{g(x+h)-g(x)}cdot{g(x+h)-g(x)}/{h}$

by the observation above,

$=lim_{t to 0}{f(g(x)+t)-f(g(x))}/{t}cdot lim_{h to 0}{g(x+h)-g(x)}/h$

by the definition of the derivative,

$=f'(g(x))cdot g'(x)$