SA: chain rule

randRange( 10, 20 ) randRange( 30, 40 ) randRange( 1, 9 ) randRange( 1, 9 ) randRange( 1, 9 ) randRange( 1, 9 ) randRange( 1, 9 ) randRange( 1, 9 ) randRange( 1, 9 ) randRange(1, 2) choice > 1 ? gprfa*fpra : fprga*gpra

This is a chain rule problem.

Given:
f(a)=fa, f(ga)=fga, f'(a)=fpra, f'(ga)=fprga,
g(a)=ga, g(fa)=gfa, g'(a)=gpra, g'(fa)=gprfa.

Let h(x)=f(g(x)).

Let h(x)=g(f(x)).

Compute h'(a).

answer

Chain rule: h'(x)=[f'(g(x))][g'(x)]

Chain rule: h'(x)=[g'(f(x))][f'(x)]

h'(a)=[f'(g(a))][g'(a)]

h'(a)=[g'(f(a))][f'(a)]

Given: g(a)=ga

Given: f(a)=fa

h'(a)=[f'(ga)][g'(a)]

h'(a)=[g'(fa)][f'(a)]

Given: f'(ga)=fprga, g'(a)=gpra

Given: g'(fa)=gprfa, f'(a)=fpra

h'(a)=[fprga][gpra]
Now you just have to do the multiplication.

h'(a)=[gprfa][fpra]
Now you just have to do the multiplication.