How I do I prove the Chain Rule for derivatives?

Question

3255 views around the world

You can reuse this answer
Creative Commons License

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)#