I know this is probably quite basic but having trouble with the following question:
Let f be a function of x and y, where x = cos(uv) and y = sin(uv).
Use the chain rule to show that:
$\displaystyle \frac{\partial f}{\partial u} = v(x\frac{\partial f}{\partial y}-y\frac{\partial f}{\partial x})$
The problem I have is I get the following:
$\displaystyle \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$ (1)
$\displaystyle \frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$ (2)
$\therefore$ from (1):
$\displaystyle \frac{\partial f}{\partial u}=\frac{\partial x}{\partial u}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial v}\frac{\partial v}{\partial x})$ (3)
&
$\displaystyle \frac{\partial f}{\partial v}=\frac{\partial y}{\partial v}(\frac{\partial f}{\partial y} - \frac{\partial f}{\partial u}\frac{\partial u}{\partial y})$ (4)
Then:
$\displaystyle \frac{\partial x}{\partial u} = -vsin(uv), \frac{\partial x}{\partial v} = -usin(uv), \frac{\partial y}{\partial u} = vcos(uv), \frac{\partial y}{\partial v} = ucos(uv)$
Plugging in the partial derivatives for x and y in to (3) I get:
$\displaystyle \frac{\partial f}{\partial u}=-vy(\frac{\partial f}{\partial x} + \frac{1}{uy}\frac{\partial f}{\partial v})$
Doing the same with (4) I get:
$\displaystyle \frac{\partial f}{\partial v}=ux(\frac{\partial f}{\partial y} - \frac{1}{vx}\frac{\partial f}{\partial u})$
After substituting for $\displaystyle \frac{\partial f}{\partial v}$ and some algebra, I get the following:
$\displaystyle \frac{\partial f}{\partial u}=-vy\frac{\partial f}{\partial x} - vx\frac{\partial f}{\partial y} + \frac{\partial f}{\partial u}$
So I'm close but not quite there. Any thoughts? What silly mistake have I made?
Thanks in advance.