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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Sign up to join this communityQuestion: 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}$
HINT
Let
$$f(x,y) = x^y=e^{y\log x}$$
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 $$ \frac{\text d}{\text dx}a^x=a^x\ln a $$