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}$
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$\begingroup$ If you have a single variable function $f(y) = a^y$, where $a$ is some constant how would you take its derivative? $\endgroup$– rubikscube09May 5, 2018 at 22:33
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2$\begingroup$ Hint: $x^y = e^{y\log x}$ <- to short for anwser v_v $\endgroup$– SK19May 5, 2018 at 22:34
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$\begingroup$ It's just the rule for exponentiation. Think about the derivative of 2^y. It is ln(2) 2^y. $\endgroup$– johnnybMay 8, 2018 at 2:42
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2 Answers
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HINT
Let
$$f(x,y) = x^y=e^{y\log x}$$
and use
$$[e^{g(y)}]'=e^{g(y)}\cdot g'(y)$$
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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 $$ \frac{\text d}{\text dx}a^x=a^x\ln a $$
