How do you calculate the derivative after a change of variables?

Question

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.

Answer 1

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.

Answer 2

More basic than the chain rule:

$f(x(\theta),y(\theta)) = \sin^2(2\theta)+\cos^2(2\theta) = 1$, so $\frac{df}{d\theta} = \frac{d}{d\theta} (1) = 0$

Answer 3

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*}