Partial Derivative of f(x,y) = x^y

Question

This is a homework problem I'm somewhat perplexed on. I thought it was straight forward, but I was incorrect in that assessment.

$$ f(x,y) = x^y\\ \text{find: }f_x(x,y)\text{ and }f_y(x,y)\\ $$ So I attempted to simplify my equation into something friendlier, but I think I botched it. Here is what I tried:

$$ \begin{align} f(x,y) = e^{ln(x^y)}\\ = e^{y*ln(x)}\\ =e^y + e^{ln(x)}\\ =e^y + x \end{align} $$

Which should have meant that: $$ f_x(x,y) = 0 + 1 = 1\\ f_y(x,y) = e^y + 0 = e^y $$

However, my answers are not correct. I'm not sure what I should have done differently, but I've always been kinda dumb when it came to exponentials. Can anyone point me in the correct direction?

Answer 1

$e^{a+b} = e^a \times e^b$, and not the other way around. Your expression just remains $e^{y \ln(x)}$. Now differentiate this.

Answer 2

This is typical case where logarithmic differentiation makes life simpler $$F=x^y$$ $$\log(F)=y \log(x)$$ So $$\frac 1F \frac{dF}{dx}=\frac yx$$ $$\frac 1F \frac{dF}{dy}=\log(x)$$ So $$\frac{dF}{dx}=\frac yx x^y=y x^{y-1}$$ $$\frac{dF}{dy}=x^y\log(x)$$