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I'm trying to show the partial derivative with respect to $x$ for $e^{xy}$ is $ye^{xy}$ using the limit definition, so $ \lim_{h \to 0} \frac{f(x + h,\ y) \ - f(x,y)}{(h)} \ = \lim_{h \to 0} \frac{e^{x+h},\ y\ -\ e^{xy},\ y}{h}$ can I then say $ \lim_{h \to 0} \frac{e^{x+h},\ -\ e^{xy}}{h}?$

I've been given the hint $= \lim_{h \to 0} \frac{e^{h}\ -\ 1}{h} = 1$

I know we can separate as so $e^{x+h} = e^x + e^h$ but I don't see how I can arrive at the correct answer.

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3 Answers

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For $y\neq 0$ we have \begin{align} \lim_{h \to 0} \frac{e^{(x+h)x} -e^{xy}}{h} &= y\lim_{h \to 0} \frac{e^{xy+hy} -e^{xy}}{hy} \\ &= y\lim_{h \to 0} \frac{e^{xy+u(h)} -e^{xy}}{u(h)}, \qquad u(h)=yh \\ &= y\lim_{u \to 0} \frac{e^{xy+u} -e^{xy}}{u}, \\ &= ye^{xy}\underbrace{\lim_{u \to 0} \frac{e^{u} -1}{u}}_{\lim_{u \to 0} \frac{e^{u} -1}{u}=1}, \\ &= ye^{xy} \\ \end{align}

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$$\lim _{ h\to 0 } \frac { f(x+h,y)-f(x,y) }{ h } =\lim _{ h\to 0 } \frac { e^{ \left( x+h \right) y }-e^{ xy } }{ h } =\lim _{ h\to 0 } \frac { e^{ xy }\left( { e }^{ yh }-1 \right) }{ hy } y={ ye }^{ xy }\\ \\ $$

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\begin{eqnarray*} \frac{ \partial f}{\partial x} =\lim_{ h \rightarrow 0} \frac{e^{(x+h)y} -e^{xy}}{h} = \lim_{ h \rightarrow 0} \frac{e^{xy}(e^{hy} -1)}{h}\underbrace{=}_{H=hy} \lim_{ H \rightarrow 0} \frac{e^{xy}(e^{H} -1)y}{H}=y e^{xy} \end{eqnarray*} This limit is easily evaluated using the hint provided in the question.

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