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.

  • 2
    $\begingroup$ 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? $\endgroup$ Oct 19 '20 at 15:46
  • $\begingroup$ What have you done? Where are the difficulties you're facing ? $\endgroup$
    – marwalix
    Oct 19 '20 at 15:46
  • $\begingroup$ Welcome to Maths SX! Use the definition of $x^y$, which is $\mathrm e^{y\ln x}$. $\endgroup$
    – Bernard
    Oct 19 '20 at 15:47
  • $\begingroup$ @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. $\endgroup$
    – Arthur
    Oct 19 '20 at 15:53
  • $\begingroup$ I added some details to the question. $\endgroup$
    – cerebrus6
    Oct 19 '20 at 15:58

if our function is:

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


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


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

  • $\begingroup$ Thank you again @mattos I should have been more careful $\endgroup$
    – Buraian
    Oct 19 '20 at 16:43
  • $\begingroup$ 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. $\endgroup$
    – cerebrus6
    Oct 19 '20 at 16:54
  • $\begingroup$ @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 $y$. $\endgroup$
    – Arthur
    Oct 20 '20 at 12:43

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