Question

If `F(x)=f(3f(4f(x)))` where f(0)=0 and f'(0)=2, how do you find F'(0)?

Answer 1

The value of `F'(0)` is `96`.

Explanation:

Consider:

`y = f(g(x))`

Then by the chain rule the derivative is given by `f'(g(x)) * g'(x)`. Thus the derivative of `y = f(g(h(x))` will be derived as being:

`f'(g(h(x)) * "derivative of "g(h(x))`

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

If we return our attention to the problem at hand, we get:

`F'(x) = f'(3f(4f(x)) * 3f'(4f(x)) * 4f'(x)`

`F'(0) = f'(3f(4f(0)) * 3f'(4f(0)) * 4f'(0)`

Substituting our knowns:

`F'(0) = f'(3f(0)) * 3f'(0) * 4(2)`

`F'(0) = f'(0) * 3(2) * 8`

`F'(0) = 2 * 6 * 8`

`F'(0) = 96`

Hopefully this helps!