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How do find partial derivative $f_x$ of $f(x,y)=\ln(x^2y)$ using definition? I know that the answer is $\frac2x$, but I can't see how to get there by using limit $$\lim_\limits{\Delta x\to 0}\frac{\ln((x+\Delta x)^2y)-\ln(x^2y)}{\Delta x}.$$

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  • $\begingroup$ You can start by using the properties of the $\ln$ function. First, $\ln((x+\Delta x)^2y)=\ln(x+\Delta x)^2+\ln y$. Do the same with $\ln (x^2y)$... $\endgroup$ – Bernard Massé Mar 12 '17 at 18:49
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Write \begin{align} L&=\frac{\ln((x+\Delta x)^2y))-\ln(x^2y)}{\Delta x} \\ &=\frac{2\ln(x+\Delta x)+\ln(y)-2\ln(x)-\ln(y)}{\Delta x}\\ &=2\left(\frac{\ln(x+\Delta x)-\ln(x)}{\Delta x}\right) \end{align}

which is recognized as the derivative of $2\ln(x)$ for $\Delta x\rightarrow 0$.

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