Error while calculating the derivative of a function to the power of another function

Question

I asked myself how one may calculate the derivative of a function that looks like this:

$h(x) = f(x)^{g(x)}$

And used the following process of derivation

$h'(x) = \frac{d}{dx}(f(x) \cdot 1^{g(x)})$

$h'(x) = f(x) \cdot \frac{d}{dx}(1^{g(x)}) + 1^{g(x)} \cdot f'(x)$

$h'(x) = f(x) \cdot 1^{g(x)} \cdot ln(1) \cdot g'(x) + 1^{g(x)} \cdot f'(x)$

$h'(x) = f'(x)^{g(x)}$

But when checking with a calculator it's clear than the result is wrong, I suspect my mistake is very obvious but I'm not able to find it. So the question is, could you tell me what I did wrong?

Answer 1

Use logarithmic differentiation if . $\, y= u^v$ where both u and v are functions of x. Take log on both sides

$\, ln(y) =vln(u)$

Now differentiate wrt x

$\, \frac1y. \frac{dy} {dx} = \frac{vu'} {u} + ln(u) v'$

Therefore $\, \frac{dy} {dx} = y(\frac{vu'} {u} + ln(u) v') $

You got it wrong when you assumed $\, f(x) ^{g(x)} = f(x). 1^{g(x)}$

Answer 2

HINT

Let use that

$$f(x)^{g(x)}=e^{g(x)\log f(x)}$$

and refer to the chain rule

$$(e^{r(x)})'=r'(x)e^{r(x)}$$

Answer 3

Hint: \begin{align*} f(x)^{g(x)}=\left(f(x)\cdot 1\right)^{g(x)}=f(x)^{g(x)}\cdot 1^{g(x)} \end{align*}