If $z=(x+y)f(ax+by)$ show that $(b(\partial z/ \partial x)-a(\partial z/ \partial y))(x-y)=(b-a)z$

Question

If $z=(x+y)f(ax+by)$ show that $(b(\partial z/ \partial x)-a(\partial z/ \partial y))(x-y)=(b-a)z$.

I don't really know where to start. I have written the total derivative for $f$ and $z$ assuming that $f=f(x,y)$ and $z=z(x,y)$ but It doesn't seem to help.

Answer 1

The notation $f(ax+by)$ implies that $f$ only takes one real number as an argument, so $f = f(x,y)$ is incorrect. $z = z(x,y)$ is correct.

First compute the partial derivatives of $z$: \begin{align*} \frac{\partial z}{ \partial x} &= f(ax+by) + (x+y)f'(ax+by)a \\ \frac{\partial z}{ \partial y} &= f(ax+by) + (x+y)f'(ax+by)b \end{align*} Now \begin{align*} \left(b\frac{\partial z}{ \partial x} -a\frac{\partial z}{ \partial y} \right) (x-y) &= \left( bf(ax+by)-af(ax+by) \right) (x-y) \\ &= (b-a)f(ax+by)(x-y) \\ &=(b-a)z \end{align*}