Question: Find the first-order partial derivatives and second-order partial derivatives for: $f(x,y) = x^y$

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

Can someone explain why $ \frac{\partial f}{\partial y} = x^{y}\log{x}$

  • $\begingroup$ If you have a single variable function $f(y) = a^y$, where $a$ is some constant how would you take its derivative? $\endgroup$ May 5, 2018 at 22:33
  • 2
    $\begingroup$ Hint: $x^y = e^{y\log x}$ <- to short for anwser v_v $\endgroup$
    – SK19
    May 5, 2018 at 22:34
  • $\begingroup$ It's just the rule for exponentiation. Think about the derivative of 2^y. It is ln(2) 2^y. $\endgroup$
    – johnnyb
    May 8, 2018 at 2:42

2 Answers 2




$$f(x,y) = x^y=e^{y\log x}$$

and use

$$[e^{g(y)}]'=e^{g(y)}\cdot g'(y)$$


For a partial derivative you hold one variable constant while you differentiate with respect to another. So $$ \frac{\partial f}{\partial y}=x^y\ln x $$ is equivalent to $$ \frac{\text d}{\text dx}a^x=a^x\ln a $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .