Question

Let r(x)=f(g(h(x))), where h(1)=2, g(2)=3, h'(1)=4, g'(2)=5, and f'(3)=6, how do you find r'(1)?

Answer 1

The value of `r'(1)` is `120`.

Explanation:

The chain rule for two functions is

`d/dx(f(g(x)) = f'(g(x)) * g'(x)`

So it would make sense if

`d/dx(f(g(h(x))) = h'(x) * g'(h(x)) * f'(g(h(x))`

We don't have any concrete functions here to work with, but we do know the values of the functions/derivatives at certain points.

`r'(x) = h'(x) * g'(h(x)) * f'(g(h(x))`

`r'(1) = h'(1) * g'(h(1)) * f'(g(h(1)))`

`r'(1) = 4 * g'(2) * f'(g(2))`

`r'(1) = 4 * 5 * f'(3)`

`r'(1) = 4 * 5 * 6`

`r'(1) = 120`

Hopefully this helps!