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This is a homework problem I'm somewhat perplexed on. I thought it was straight forward, but I was incorrect in that assessment.

$$ f(x,y) = x^y\\ \text{find: }f_x(x,y)\text{ and }f_y(x,y)\\ $$ So I attempted to simplify my equation into something friendlier, but I think I botched it. Here is what I tried:

$$ \begin{align} f(x,y) = e^{ln(x^y)}\\ = e^{y*ln(x)}\\ =e^y + e^{ln(x)}\\ =e^y + x \end{align} $$

Which should have meant that: $$ f_x(x,y) = 0 + 1 = 1\\ f_y(x,y) = e^y + 0 = e^y $$

However, my answers are not correct. I'm not sure what I should have done differently, but I've always been kinda dumb when it came to exponentials. Can anyone point me in the correct direction?

Niko
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2 Answers2

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$e^{a+b} = e^a \times e^b$, and not the other way around. Your expression just remains $e^{y \ln(x)}$. Now differentiate this.

  • So, differentiating $e^{y \ln(x)}$ was the correct move. It yielded $f_x(x,y) = e^{y \ln(x)} * (y/x)$ and $f_y(x,y) = e^{y \ln(x)} * ln(x)$, but I'm not sure I understand why it can't be "the other way around". Can you elaborate on that point? – Niko Jun 26 '15 at 03:15
  • This is one of the basic rules of exponentiation. "When we have the same base, we add the powers". As for why it couldn't be the other way around, look at this for instance: If it were, then $2^3 = 2^{1 \times 3} = 2^1 + 2^3$, which is absurd. –  Jun 26 '15 at 03:22
  • You're right, that is absurd. Thank's for pointing that out. I don't know where my brain's been, but that made everything make sense. Thanks! – Niko Jun 26 '15 at 06:54
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This is typical case where logarithmic differentiation makes life simpler $$F=x^y$$ $$\log(F)=y \log(x)$$ So $$\frac 1F \frac{dF}{dx}=\frac yx$$ $$\frac 1F \frac{dF}{dy}=\log(x)$$ So $$\frac{dF}{dx}=\frac yx x^y=y x^{y-1}$$ $$\frac{dF}{dy}=x^y\log(x)$$

Claude Leibovici
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