the partial derivative of $f$ in terms of $g$ and $h$ of $f(x,y)=g(x)^{h(y)}$

Question

Find the partial derivative of $f$ in terms of $g$ and $h$ of $f(x,y)=g(x)^{h(y)}$

I want to show it using the limit definition.

My attempt:

$$\partial_1f(g,h)=\lim\limits_{k \to 0}\frac{f(g+k,h)-f(g,h)}{k}$$ $$=\lim\limits_{k \to 0}\frac{g(x+k)^{h(y)}-g(x)h(y)}{k}$$

There is where I am getting stuck.

Likewise,

$$\partial_2f(g,h)=\lim\limits_{k \to 0}\frac{f(g,h+k)-f(g,h)}{k}$$ $$=\lim\limits_{k \to 0}\frac{g(x)^{h(y+k)}-g(x)h(y)}{k}$$ $$=g(x)\lim\limits_{k \to 0}\frac{g(x)^{h(y)}g(x)^{h(k)}-h(y)}{k}$$

Also stuck here..

Answer 1

$f(x,y) = e^{h(y)\ln g(x)}\\ \frac {\partial f}{\partial x} = e^{h(y)\ln g(x)} (\frac {h}{g})\frac{\partial g}{\partial x}\\ \frac {\partial f}{\partial y} = e^{h(y)\ln g(x)} ({\ln g}) \frac{\partial h}{\partial y}$