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
Final Answer
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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