Partial derivative of a composite function

How do I calculate the partial derivative of this composite function $$f(x,y)=\varphi (\frac yx,x^2-y^2,y-x)$$ I don't know the steps I have to take to solve this. Thanks!

Solution 1:

If we identify the functions of $ \ x \ $ and $ \ y \ $ involved in the definition of the function $ \ \varphi \ $ by

$$f(x,y) \ = \ \varphi (\ \underbrace{\frac yx}_u \ , \ \underbrace{ x^2-y^2}_v \ , \ \underbrace{y-x}_w \ ) \ , $$

we can use the multivariate extension of the Chain Rule to write

$$ \frac{\partial f}{\partial x} \ = \ \frac{\partial \varphi}{\partial u}\frac{\partial u}{\partial x} \ + \ \frac{\partial \varphi}{\partial v}\frac{\partial v}{\partial x} \ + \ \frac{\partial \varphi}{\partial w}\frac{\partial w}{\partial x} \ \ , $$

$$ \frac{\partial f}{\partial y} \ = \ \frac{\partial \varphi}{\partial u}\frac{\partial u}{\partial y} \ + \ \frac{\partial \varphi}{\partial v}\frac{\partial v}{\partial y} \ + \ \frac{\partial \varphi}{\partial w}\frac{\partial w}{\partial y} \ \ .$$

We can find from the available information that

$$ \frac{\partial u}{\partial x} \ = \ -\frac{y}{x^2} \ , \ \frac{\partial v}{\partial x} \ = \ 2x \ \ , \ \frac{\partial w}{\partial x} \ = \ -1 \ \ , $$

$$ \frac{\partial u}{\partial y} \ = \ \frac{1}{x} \ , \ \frac{\partial v}{\partial y} \ = \ -2y \ \ , \ \frac{\partial w}{\partial y} \ = \ 1 \ \ . $$

However, since we know nothing else about the function $ \ \varphi (u,v,w) \ , $ we cannot develop the partial derivatives for $ \ f \ $ any further.