1
$\begingroup$

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.

$\endgroup$
10
  • 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
1
$\begingroup$

if our function is:

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

Then,

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

And,

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

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.