On derivatives of composite and inverse functions

Question

Question

Suppose $g$ is the inverse of a one-to-one differentiable function $f$ and $H = g \circ g\ $. If $f(4) = 3\ $, $g(4) = 5\ $, $f'(4) = \frac 1 2\ $ and $f'(5) = 2\ $, find $H'(3)\ $.

My working

$\because H(x) = g^2(x)$

$\therefore H'(x) = 2g(x)g'(x)$

$\because g'(x) = \frac 1 {f'[f^{-1}(x)]}$

$\therefore H'(x) = \frac {2f^{-1}(x)} {f'[f^{-1}(x)]}$

$\because f(4) = 3$

$\therefore f^{-1}(3) = 4$

$\implies H'(3) = \frac {2(4)} {f'(4)} = \frac {2(4)} {\frac 1 2} = 16$

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However, the answer given is $H'(3) = 1\ $. I am not sure where I have gone wrong (although I have a hunch it is probably in my derivation of $H'(x)$). Any help will be greatly appreciated :)

Answer 1

You went wrong right at the beginning (assuming that you wrote the problem correctly).

$$H = g \circ g = g(g(x)) $$

Composition not multiplication. I'm assuming that you can get $g(3)= 4$ easily.

So $$H' = g \circ g = g'(g(x))g'(x)$$

Using rule for derivate inverses.

$$g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(4)} = 2$$ $$g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(5)} = \frac{1}{2}$$