How do you calculate the derivative after a change of variables? How would you calculate $df \over dθ$ if $f(x,y) = x^2+y^2$ where $x = \sin 2θ$ and $y = \cos 2θ$?
I tried Wolfram and using the product rule but I can't seem to get anywhere.
 A: More basic than the chain rule:
$f(x(\theta),y(\theta)) = \sin^2(2\theta)+\cos^2(2\theta) = 1$, so $\frac{df}{d\theta} = \frac{d}{d\theta} (1) = 0$
A: Chain rule:
$$
\frac{\partial f}{\partial\theta} = \frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial\theta} + \frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial\theta}.
$$
PS: The question qas edited after I posted this, stating that $f(x,y)=x^2+y^2$.  That of course makes it much simpler than if a different function had been involved, since one of the Pythagorean identities of trigonometry simplifies it.
A: Here is a worked through answer using the the chain rule, although I would suggest Steven's approach is best:
$$
\frac{df}{d\theta} = \frac{\partial f}{\partial x}\frac{dx}{d\theta} + \frac{\partial f}{\partial y}\frac{dy}{d\theta}
$$
So
$$
\frac{\partial f}{\partial x} = 2x \quad \frac{\partial f}{\partial y} = 2y
$$
$$
\frac{dx}{d\theta} = 2cos 2\theta \quad \frac{y}{d\theta} = -2sin2\theta
$$
\begin{align*}
   \frac{df}{d\theta} & =  4xcos2\theta - 4y sin2\theta\\
   & =  4sin2\theta cos2\theta - 4cos2\theta sin2\theta \\
   & = 0
\end{align*}
