Differentiating F(u, v) = f(x(u, v), y(u, v))

Question

Let $f(x,y)$ be a given differentiable function.

Consider the function $F(u,v) = f( x(u,v), y(u,v) )$ where

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

Prove that

$u^3\frac{dF}{du} - \frac{dF}{dv} = -2\sqrt y\frac{df}{dy}$

I'm having difficulty differentiating this function. I think it uses the chain rule but I'm unsure of how to go about it. Any help is very much appreciated, thanks!

Answer 1

By the chain rule,

$$F_u(u,v)=f_xx_u+f_yy_u$$

and

$$F_v(u,v)=f_xx_v+f_yy_v$$

Now, $x_u=-1/u^3$, $x_v=-1$, $y_u=0$ and $y_v=2v$. Therefore

$$F_u(u,v)=-\frac{f_x}{u^3}\quad \iff \quad f_x=-u^3F_u$$

and

$$F_v(u,v)=-f_x+2vf_y.$$

Substituting for $f_x$ in the second equation gives

$$F_v=u^3F_u+2vf_y\quad\iff\quad u^3F_u-F_v=-2vf_y$$

Since $v=\pm\sqrt{y}$, we get

$$u^3F_u-F_v=\pm2\sqrt{y}f_y.$$