Chain Rule

Calculus Level 4

Functions f f and g g are such that: f ( x , y ) = g ( u , v ) \large f(x,y)=g(u,v) where u = x + y x y u=\dfrac{x+y}{xy} and v = x − y x y v=\dfrac{x-y}{xy} . Find the value of: x 2 ∂ f ∂ x + y 2 ∂ f ∂ y + 2 ∂ g ∂ u \large x^2\frac{\partial f}{\partial x}+y^2\frac{\partial f}{\partial y}+2\frac{\partial g}{\partial u}


The answer is 0.

Hjalmar Orellana Soto by Hjalmar Orellana Soto , 24, Chile

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2 solutions

Using chain rule we have: ∂ f ∂ x = ∂ g ∂ u ⋅ ∂ u ∂ x + ∂ g ∂ v ⋅ ∂ v ∂ x = ∂ g ∂ u ⋅ ∂ ∂ x ( x + y x y ) + ∂ g ∂ v ⋅ ∂ g ∂ x ( x − y x y ) = ∂ g ∂ u ⋅ − 1 x 2 + ∂ g ∂ v ⋅ − 1 x 2 \frac{\partial f}{\partial x}=\frac{\partial g}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial g}{\partial v}\cdot \frac{\partial v}{\partial x}=\frac{\partial g}{\partial u}\cdot \frac{\partial}{\partial x}(\frac{x+y}{xy})+\frac{\partial g}{\partial v} \cdot \frac{\partial g}{\partial x}(\frac{x-y}{xy})=\frac{\partial g}{\partial u}\cdot \frac{-1}{x^2} +\frac{\partial g}{\partial v} \cdot \frac{-1}{x^2} The same way: ∂ f ∂ y = ∂ g ∂ u ⋅ − 1 y 2 + ∂ g ∂ v ⋅ 1 y 2 \frac{\partial f}{\partial y}=\frac{\partial g}{\partial u} \cdot \frac{-1}{y^2}+\frac{\partial g}{\partial v} \cdot \frac{1}{y^2} So we can replace: x 2 ∂ f ∂ x + y 2 ∂ f ∂ y + 2 ∂ g ∂ u = − ∂ g ∂ u − ∂ g ∂ v − ∂ g ∂ u + ∂ g ∂ v + 2 ∂ g ∂ u = 0 x^2\frac{\partial f}{\partial x}+y^2\frac{\partial f}{\partial y}+2\frac{\partial g}{\partial u} =-\frac{\partial g}{\partial u}-\frac{\partial g}{\partial v}-\frac{\partial g}{\partial u}+\frac{\partial g}{\partial v}+2\frac{\partial g}{\partial u}=0

3 Helpful 0 Interesting 0 Brilliant 0 Confused

X = x 2 ∂ f ∂ x + y 2 ∂ f ∂ y + 2 ∂ g ∂ u = x 2 ( ∂ g ∂ u ⋅ ∂ u ∂ x + ∂ g ∂ v ⋅ ∂ v ∂ x ) + y 2 ( ∂ g ∂ u ⋅ ∂ u ∂ y + ∂ g ∂ v ⋅ ∂ v ∂ y ) + 2 ∂ g ∂ u = x 2 ( ∂ g ∂ u ⋅ ∂ ∂ x ( 1 y + 1 x ) + ∂ g ∂ v ⋅ ∂ ∂ x ( 1 y − 1 x ) ) + y 2 ( ∂ g ∂ u ⋅ ∂ ∂ y ( 1 y + 1 x ) + ∂ g ∂ v ⋅ ∂ ∂ y ( 1 y − 1 x ) ) + 2 ∂ g ∂ u = x 2 ( ∂ g ∂ u ( − 1 x 2 ) + ∂ g ∂ v ( 1 x 2 ) ) + y 2 ( ∂ g ∂ u ( − 1 y 2 ) + ∂ g ∂ v ( − 1 y 2 ) ) + 2 ∂ g ∂ u = − ∂ g ∂ u + ∂ g ∂ v − ∂ g ∂ u − ∂ g ∂ v + 2 ∂ g ∂ u = 0 \begin{aligned} X & = x^2 \frac {\partial f}{\partial x} + y^2 \frac {\partial f}{\partial y} + 2 \frac {\partial g}{\partial u} \\ & = x^2 \left(\frac {\partial g}{\partial u} \cdot \frac {\partial u}{\partial x} + \frac {\partial g}{\partial v} \cdot \frac {\partial v}{\partial x} \right) + y^2 \left(\frac {\partial g}{\partial u} \cdot \frac {\partial u}{\partial y} + \frac {\partial g}{\partial v} \cdot \frac {\partial v}{\partial y} \right) + 2 \frac {\partial g}{\partial u} \\ & = x^2 \left(\frac {\partial g}{\partial u} \cdot \frac {\partial}{\partial x}\left(\frac 1y+\frac 1x\right) + \frac {\partial g}{\partial v} \cdot \frac {\partial}{\partial x}\left(\frac 1y-\frac 1x\right) \right) + y^2 \left(\frac {\partial g}{\partial u} \cdot \frac {\partial}{\partial y}\left(\frac 1y+\frac 1x\right) + \frac {\partial g}{\partial v} \cdot \frac {\partial}{\partial y}\left(\frac 1y-\frac 1x\right) \right) + 2 \frac {\partial g}{\partial u} \\ & = x^2 \left(\frac {\partial g}{\partial u} \left(-\frac 1{x^2} \right) + \frac {\partial g}{\partial v} \left(\frac 1{x^2}\right) \right) + y^2 \left(\frac {\partial g}{\partial u} \left(-\frac 1{y^2}\right) + \frac {\partial g}{\partial v} \left(-\frac 1{y^2}\right) \right) + 2 \frac {\partial g}{\partial u} \\ & = - \frac {\partial g}{\partial u} + \frac {\partial g}{\partial v} - \frac {\partial g}{\partial u} - \frac {\partial g}{\partial v} + 2 \frac {\partial g}{\partial u} \\ & = \boxed{0} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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