What are the First Partial Derivatives of $x^y$

The problem I have are from the first partial derivatives of $$f(x,y) = x^y$$
What is its $$f_x(x,y)$$ and $$f_y(x,y)$$?

I need to find the critical points of $$f(x,y) = x^y + 4xy - 2y^2 + 5$$, but the $$x^y$$ is making me confused.
The answer I get when I try to find its partial derivatives are
$$f_x(x,y) = yx^{y-1}+4y$$
$$f_y(x,y) = x^y\ln x+4x-4y$$
I am stuck in this step and I am not sure if my partial derivatives are right.

• Do you want to verify an answer that you have? Also, have you seen the formula for the derivatives of the functions $f(x) = x^n$ and $f(x) = a^x$, where $n,a$ are fixed? Oct 19 '20 at 15:46
• What have you done? Where are the difficulties you're facing ? Oct 19 '20 at 15:46
• Welcome to Maths SX! Use the definition of $x^y$, which is $\mathrm e^{y\ln x}$. Oct 19 '20 at 15:47
• @Bernard That's not necessarily the definition. There are a few other common options. But I agree that that might be the easiest way to reach a solution. Oct 19 '20 at 15:53
• I added some details to the question. Oct 19 '20 at 15:58

if our function is:

$$f(x,y) = x^y$$

Then,

$$\frac{ \partial f}{\partial x} = y x^{y-1}$$

And,

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

• Thank you again @mattos I should have been more careful Oct 19 '20 at 16:43
• Ah, so that's why I've been going in circles since earlier and can't seem to equate $x^y$ and $e^{x ln y}$ no matter what method I use. Haha. Thank you also @mattos and Buraian. Oct 19 '20 at 16:54
• @cerebrus6 We have $x^y = e^{\ln(x^y)}$, by the definition of logarithm, and $e^{\ln(x^y)} = e^{x\ln y}$ by standard logarithm properties. Or it could be that $x^y = e^{y\ln x}$ is how you define $x^y$ for real positive $x$ and real $y$. Oct 20 '20 at 12:43