# A question about chain rule for partial derivatives

I know this isn't supposed to be difficult, but I'm not sure how to use the chain rule.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable and define $u: \mathbb{R}^2 \rightarrow \mathbb{R}$ by $u(x,y):=e^{2x}f(ye^{-x})$. I want to show that $u$ satisfies the partial differential equation $$\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}=2u.$$

I know how to take partial derivatives with respect to $x$ and $y$, but I don't know how to implement the chain rule here. How do I do this? I appreciate your help.

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Do you know how to use the chain rule to find $(g\circ h)'$? – Git Gud May 31 '13 at 19:24
$(g\circ h)'(x)=g'(h(x))h'(x)$ – user4167 May 31 '13 at 19:29
Prove the following equation via partial differentiations of $u(x,y)$ by Git Gud's hint. – NasuSama May 31 '13 at 19:30

## 1 Answer

Here is how one starts off proving the equality.

Consider differentiating the function with respect to certain variable... We have:

$$\frac{\partial u}{\partial x} = 2e^{2x}f(ye^{-x}) + (-ye^{-x}e^{2x}f'(ye^{-x}))$$ $$= 2e^{2x}f(ye^{-x}) - ye^{x}f'(ye^{-x})$$ $$\frac{\partial u}{\partial y} = e^{-x}e^{2x}f'(ye^{-x})$$

Substitute the given and the following equations for the given equality, which yields:

$$LHS = 2e^{2x}f(ye^{-x}) - ye^{x}f'(ye^{-x}) + y(e^{-x}e^{2x}f'(ye^{-x}))$$ $$= 2e^{2x}f(ye^{-x}) - ye^{x}f'(ye^{-x}) + ye^{x}f'(ye^{-x})$$ $$= 2e^{2x}f(ye^{-x})$$ $$= 2(e^{2x}f(ye^{-x}))$$ $$= 2u = RHS$$

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