# Multivariable Chain Rule – Calculating partial derivatives – Exercise 6489

Exercise

Given the differentiable functions

$$z=x^2\ln y$$

$$y=3u-2v$$

$$x=\frac{u}{v}$$

Calculate the partial derivatives

$$z'_u, z'_v$$

$$z'_u=\frac{u}{v^2}(2\ln (3u-2v)+\frac{3u}{3u-2v})$$

$$z'_v=\frac{u^2}{v^2}(\frac{-2v}{v^2}\ln (3u-2v)-\frac{2}{3u-2v})$$

Solution

Put the functions x and y in function z and get

$$z=x^2\ln y=$$

$$={(\frac{u}{v})}^2\ln (3u-2v)$$

Calculate the partial derivatives of z.

$$z'_u=\frac{2u}{v^2}\ln (3u-2v)+\frac{u^2}{v^2}\cdot\frac{1}{3u-2v}\cdot 3=$$

$$=\frac{u}{v^2}(2\ln (3u-2v)+\frac{3u}{3u-2v})$$

$$z'_v=\frac{-2v\cdot u^2}{v^4}\ln (3u-2v)+\frac{u^2}{v^2}\cdot\frac{1}{3u-2v}\cdot (-2)=$$

$$=\frac{u^2}{v^2}(\frac{-2v}{v^2}\ln (3u-2v)-\frac{2}{3u-2v})$$

Note: one can calculate the derivatives directly using the chain rule.

$$z'_u=z'_x\cdot x'_u + z'_y\cdot y'_u$$

$$z'_v=z'_x\cdot x'_v + z'_y\cdot y'_v$$

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