Exercise
Given the differentiable functions
z = x 2 ln y z=x^2\ln y z = x 2 ln y
y = 3 u − 2 v y=3u-2v y = 3 u − 2 v
x = u v x=\frac{u}{v} x = v u
Calculate the partial derivatives
z u ′ , z v ′ z'_u, z'_v z u ′ , z v ′
Final Answer
Show final answer
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 = z=x^2\ln y= z = x 2 ln y =
= ( u v ) 2 ln ( 3 u − 2 v ) ={(\frac{u}{v})}^2\ln (3u-2v) = ( v u ) 2 ln ( 3 u − 2 v )
Calculate the partial derivatives of z.
z u ′ = 2 u v 2 ln ( 3 u − 2 v ) + u 2 v 2 ⋅ 1 3 u − 2 v ⋅ 3 = z'_u=\frac{2u}{v^2}\ln (3u-2v)+\frac{u^2}{v^2}\cdot\frac{1}{3u-2v}\cdot 3= z u ′ = v 2 2 u ln ( 3 u − 2 v ) + v 2 u 2 ⋅ 3 u − 2 v 1 ⋅ 3 =
= u v 2 ( 2 ln ( 3 u − 2 v ) + 3 u 3 u − 2 v ) =\frac{u}{v^2}(2\ln (3u-2v)+\frac{3u}{3u-2v}) = v 2 u ( 2 ln ( 3 u − 2 v ) + 3 u − 2 v 3 u )
z v ′ = − 2 v ⋅ u 2 v 4 ln ( 3 u − 2 v ) + u 2 v 2 ⋅ 1 3 u − 2 v ⋅ ( − 2 ) = z'_v=\frac{-2v\cdot u^2}{v^4}\ln (3u-2v)+\frac{u^2}{v^2}\cdot\frac{1}{3u-2v}\cdot (-2)= z v ′ = v 4 − 2 v ⋅ u 2 ln ( 3 u − 2 v ) + v 2 u 2 ⋅ 3 u − 2 v 1 ⋅ ( − 2 ) =
= u 2 v 2 ( − 2 v v 2 ln ( 3 u − 2 v ) − 2 3 u − 2 v ) =\frac{u^2}{v^2}(\frac{-2v}{v^2}\ln (3u-2v)-\frac{2}{3u-2v}) = v 2 u 2 ( v 2 − 2 v ln ( 3 u − 2 v ) − 3 u − 2 v 2 )
Note: one can calculate the derivatives directly using the chain rule.
z u ′ = z x ′ ⋅ x u ′ + z y ′ ⋅ y u ′ z'_u=z'_x\cdot x'_u + z'_y\cdot y'_u z u ′ = z x ′ ⋅ x u ′ + z y ′ ⋅ y u ′
z v ′ = z x ′ ⋅ x v ′ + z y ′ ⋅ y v ′ z'_v=z'_x\cdot x'_v + z'_y\cdot y'_v z v ′ = z x ′ ⋅ x v ′ + z y ′ ⋅ y v ′
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