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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.

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  • $\begingroup$ Your $x_\theta$ and $y_\theta$ terms don't make sense. Why did you put $32r\cos\theta$, etc? $\endgroup$ Oct 3, 2020 at 19:11
  • $\begingroup$ @NinadMunshi I got rid of them too but the answer is still wrong $\endgroup$
    – Si Random
    Oct 3, 2020 at 19:14
  • $\begingroup$ Except getting rid of them is not correct. What is the formula for chain rule in 2 variables? $\endgroup$ Oct 3, 2020 at 19:15

2 Answers 2

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It is easier if you substitute $x$ and $y$ directly $$g(r,\theta)=\frac{1}{x+y^2}=\frac{1}{32r\cos(\theta)+9r^2\sin^2(\theta)}$$

So $$\frac{\partial g}{\partial \theta}=-\frac{-32r\sin(\theta)+18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$


Or $$\frac{\partial g}{\partial \theta}=\frac{\partial g }{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g }{\partial y}\frac{\partial y}{\partial \theta}$$ $$=-\frac{1}{(x+y^2)^2}(-32r\sin(\theta)+2y(3r\cos(\theta)))$$ $$=\frac{32r\sin(\theta)-18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$

Then evaluating at $(r,\theta)=(2\sqrt{2},\frac{\pi}{4})$ you should get $-\frac{1}{1250}$.

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    $\begingroup$ OMG ares <333333 thank you $\endgroup$
    – Si Random
    Oct 3, 2020 at 19:18
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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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  • $\begingroup$ Woah that's interesting. whered you get 64 from? thanks again btw! $\endgroup$
    – Si Random
    Oct 3, 2020 at 19:18
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    $\begingroup$ @DoctorReality We have $$x=32r\cos \theta = 32\cdot 2\sqrt 2 \cdot \frac 1{\sqrt 2}=64$$ $\endgroup$
    – user
    Oct 3, 2020 at 19:19
  • $\begingroup$ OOOOOOOOOOOOOOOOH! I shoulda done that $\endgroup$
    – Si Random
    Oct 3, 2020 at 19:20
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    $\begingroup$ @DoctorReality We don't need to find a general expression for the derivative, after the first step it suffices to plug in the values for the given point. $\endgroup$
    – user
    Oct 3, 2020 at 19:21

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