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Find the partial derivative in respect to $y$ of $f(x,y,z)=x^y$ using the limit definition.

My attempt:

$$\lim\limits_{h \to 0} \frac{f(x,y+h,z)-f(x,y,z)}{h}$$ $$\lim\limits_{h \to 0} \frac{x^{y+h}-x^y}{h}$$ $$\lim\limits_{h \to 0} \frac{x^yx^h-x^y}{h}$$

Now I am stuck because I don't know how to apply my log rules to this.

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  • $\begingroup$ Well, if you know it's a "log rule" (which I interpret as: "I know I should remember it from a very specific moment of my tuition history, but I don't") perhaps you'd better look it up in the high school book, no? $\endgroup$ – user228113 Feb 11 '17 at 17:49
  • $\begingroup$ Your comment does not help me in any way,shape,or form. Thanks $\endgroup$ – combo student Feb 11 '17 at 18:08
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A few more steps:

$$\begin{align}\lim_{h\to0}\frac{x^yx^h-x^y}h&=x^y\lim_{h\to0}\frac{x^h-1}h\\&=x^y\lim_{h\to0}\frac{e^{h\ln(x)}-1}h\end{align}$$

Let $h\ln(x)=u$,

$$=x^y\lim_{u\to0}\frac{e^u-1}{\frac u{\ln(x)}}=x^y\ln(x)\lim_{u\to0}\frac{e^u-1}u$$

$$f_y=x^y\ln(x)$$

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  • $\begingroup$ why can we factor out an $x^y$ outside the limit? $\endgroup$ – combo student Feb 11 '17 at 17:47
  • $\begingroup$ @combostudent Because it is independent of $h$?? $\endgroup$ – Simply Beautiful Art Feb 11 '17 at 17:48
  • $\begingroup$ another question, if $u=hln(x)$ why can you write $u \rightarrow 0$? $\endgroup$ – combo student Feb 11 '17 at 17:58
  • $\begingroup$ @combostudent if $h\to0$, what can you conclude about $u\to?$, assume $x>0$. $\endgroup$ – Simply Beautiful Art Feb 11 '17 at 18:21

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