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 $$\lim_\limits{\Delta x\to 0}\frac{\ln((x+\Delta x)^2y)-\ln(x^2y)}{\Delta x}.$$
2026-04-12 18:57:08.1776020228
Finding partial derivative $f_x$ of $\ln(x^2y)$ using definition.
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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 $\Delta x\rightarrow 0$.