Cant understand how chain rule works

Question

Let $w(x,y)$ be a function of class $C^2$ in the variables $x$ and $y$, and let $x=u+v$, $y=u-v$, show that: \begin{align} \frac{\partial^2 w}{\partial u \partial v} = \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2}\end{align}

My attempt:

What we are looking for is $\frac{\partial}{\partial u}(\frac{\partial w}{\partial v})$, so, by the chain rule:

$$\frac{\partial w}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} = \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}$$ and similarly for $\frac{\partial w}{\partial u}$.

Now, what we need is: $$\frac{\partial}{\partial u}(\frac{\partial w}{\partial v}) = \frac{\partial}{\partial u}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}) = \frac{\partial}{\partial u}(\frac{\partial f}{\partial x}) - \frac{\partial}{\partial u}(\frac{\partial f}{\partial y})$$

But I can’t seem to grasp what $\frac{\partial}{\partial u}(\frac{\partial f}{\partial x})$ is or how am I supposed to apply the chain rule again in this case.

Answer 1

To elaborate on NDewolf's answer, apply the same trick in your first step to the term $\partial/\partial u$. That is, write $$ \frac{\partial}{\partial u} = \frac{\partial}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial}{\partial y}\frac{\partial y}{\partial u}, $$ and use the fact that $$ \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y}\right)(g) = \frac{\partial}{\partial x}(g) + \frac{\partial}{\partial y}(g) $$ where $g$ is any function (ie, it could be the derivative of a function).

Answer 2

Try to apply the same trick as in your first step. Use the chain rule to differentiate the partial derivatives $\frac{\partial w}{\partial x}$, $\frac{\partial w}{\partial y}$ with respect to $x$ and $y$.

Answer 3

First of all you have to note that you rather have $w(u,v)=f(x,y)$ and in addition $x=x(u,y)=u+v, \ y=y(u,v)=u-v$. Then everything follows.

$\frac{∂w}{∂v}=\frac{∂f}{∂x}\frac{∂x}{∂v}+\frac{∂f}{∂y}\frac{∂y}{∂v}=\frac{∂f}{∂x}-\frac{∂f}{∂y}$

$\frac{∂^2w}{∂u ∂v}=\frac{∂}{∂u}(\frac{∂w}{∂v})=\frac{∂}{∂u}(\frac{∂f}{∂x}-\frac{∂f}{∂y})=(\frac{∂^2f}{∂x^2}\frac{∂x}{∂u}+\frac{∂^2f}{∂y ∂x}\frac{∂y}{∂u})-(\frac{∂^2f}{∂x ∂y}\frac{∂x}{∂u}+\frac{∂^2f}{∂y^2}\frac{∂y}{∂u})=\frac{∂^2f}{∂x^2}+\frac{∂^2f}{∂y ∂x}-\frac{∂^2f}{∂x ∂y}-\frac{∂^2f}{∂y^2}=\frac{∂^2f}{∂x^2}-\frac{∂^2f}{∂y^2}$

(The last equal sign is because of $C^2$, that is the order of the partial derivations is interchangeable.)

--- rk

Answer 4

You found $w_v=f_x-f_y$ and $(w_v)_u=(f_x)_u-(f_y)_u$. Note: $$\begin{align}(w_v)_u=&(f_{xx}\cdot x_u+f_{xy}\cdot y_u)-(f_{yx}\cdot x_u+f_{yy}\cdot y_u)=\\ =&(f_{xx}+f_{xy})-(f_{yx}+f_{yy})=\\ =&f_{xx}-f_{yy}. \end{align}$$