I have been worked out a solution for the following problem, but I am wondering if there is an easier way to solve it. I would be very grateful for any suggestions!
Let $z=f(x,y)$ a function of two independent variables $x$. and $y$. Show that, if we perform a change of variables with $u$ and $v$ $u=x^2+y^2$ and $u=e^{\frac{y}{x}}$, then $$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 2v \frac{\partial z}{\partial v} \:.$$
My solution: By the multivariate chain rule, we have $$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v} \:.$$ Hence it is sufficient to show that $\frac{\partial x}{\partial v} = \frac{x}{2v}$ and $\frac{\partial y}{\partial v} = \frac{y}{2v}$. There are two ways in which we can do this (I will consider only $\frac{\partial x}{\partial v}$):
- Either we calculate the partial derivatives$\frac{\partial v}{\partial x}$, $\frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ and use the fact that, by the inverse function rule, $$ \begin{bmatrix} \frac{\partial x}{\partial v} & \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial v} & \frac{\partial y}{\partial u} \end{bmatrix} = \begin{bmatrix} \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \end{bmatrix}^{-1} $$ to obtain $$\frac{\partial x}{\partial v} = \frac{\frac{\partial u}{\partial y}}{\frac{\partial v}{\partial x} \cdot \frac{\partial u}{\partial y} - \frac{\partial v}{\partial y} \frac{\partial u}{\partial x}} = \frac{x}{2v} .$$
- Or we simply notice that $\ln(u) = \frac{y}{x}$ and thus $y = x \cdot \ln(u)$. Substituting this into $v=x^2+y^2$, we get $x^2=\frac{v}{1+[\ln(u)]^2}$ and, if we partially differentiate both sides with respect to $v$, we obtain $2x\frac{\partial x}{\partial v} = \frac{1}{1+[\ln(u)]^2} = \frac{1}{1+\left[\frac{y}{x}\right]^2} = \frac{x^2}{x^2+y^2} = \frac{x^2}{v}$, as required.
