<s>[INST] Evaluating $\frac{dg}{dθ}$ at $(r,θ)=(2\sqrt{2},\frac{π}{4})$ where $g(x,y)=\frac1{x+y^2}$ using chain rule? 
Okay so my first step is to find the partial derivatives:
$$\frac{\partial \:}{\partial \:x}\left(\frac{1}{x+y^2}\right)=-\frac{1}{\left(x+y^2\right)^2}$$
$$\frac{\partial \:}{\partial \:y}\left(\frac{1}{x+y^2}\right)=-\frac{2y}{\left(x+y^2\right)^2}$$
Then I multiply each partial derivative with its respective equal and add it together:
$$(fx)(32r\cos(\theta)) + (fy)(3r(\sin(\theta))$$
Then I sub in X and Y in the partial derivatives. And then plug in $r$ and $\theta$:
$$-\left(\frac{1}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi }{4}\right)\right)^2\right)^2}\right)\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)\right)+-\left(\frac{2\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:\:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)^2\right)^2}\right)\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)$$
And I show my answer in the screenshot above. But for some reason the software is telling me it's wrong. I've also tried not multiplying the partial derivatives by the $32rcos(\theta)$ thing but it's still giving me the wrong answer.
What am I doing wrong? Thank you.
 [/INST] We have that
$$\frac{\partial g}{\partial \theta}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial \theta}=-\frac{1}{\left(x+y^2\right)^2}(-32 r \sin \theta)-\frac{2y}{\left(x+y^2\right)^2}(3r\cos \theta)=$$
with $x=64$ and $y=6$ then
$$=-\frac{1}{100^2}(-64)-\frac{12}{100^2}(6)=-\frac{8}{100^2}$$
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