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

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Answer 1 · 2017-11-28T23:08:00

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

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!

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