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$F:\mathbb{R}^2\rightarrow\mathbb{R}$ is a $C^2$ function with $F_x(1,1)=F_{yy}(1,1)=1$ and $F_y(1,1)=F_{xx}(1,1)=0$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $g(r,\theta)=F(r\cos\theta,r\sin\theta)$. I'm asked to find the value of $g_{r\theta}(\sqrt2,\pi/4)$.

This what I understand about the chain rule and what I've done so far.

I can see $g$ as the composition of $F$ and $M=(r\cos\theta,r\sin\theta)$. $M$ is differentiable at $(\sqrt2,\pi/4)$ and $F$ is differentiable at $M(\sqrt2,\pi/4)=(1,1)$ because its partial derivatives exist and are continuous. The chain rule tells me that (*) $\frac{\partial g}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta$. Then I have to take the derivative of that with respect to $\theta$, and I don't know how to do that.

EDIT: This is a silly question, but I'd also like to know what are the points of evaluation in (*) and/or the chain rule thesis in general. My textbook says it's understood, but I'm not too good at this.

English is not my first language, so feel free to correct any mistake.

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    $\begingroup$ Those should be partial derivatives. You can get the necessary symbol with \partial. I’ve edited it for you. $\endgroup$
    – amd
    Commented Jan 26, 2016 at 22:29
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    $\begingroup$ "This is a silly question, but I'd also like to know what are the points of evaluation in (*) and/or the chain rule thesis in general". This is far from silly and it's exactly why the usual notation for the chain rule is awful. This comment should shed some light on this. $\endgroup$
    – Git Gud
    Commented Jan 26, 2016 at 23:33

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Well, if $$ \frac{\partial g}{\partial r} = \frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta $$

Differentiating with respect to $\theta$, $$ \frac{\partial^2 g}{\partial \theta \partial r} = \frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta\right) $$

I'll do the first term: \begin{align} \frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\cos\theta\right) &= \cos\theta\frac{\partial}{\partial \theta} \left(\frac{\partial f}{\partial x}\right) + \frac{\partial (\cos \theta)}{\partial \theta} \frac{\partial f}{\partial x} \\ &= \cos\theta\frac{\partial}{\partial \theta} \left(\frac{\partial f}{\partial x}\right) - \sin \theta\frac{\partial f}{\partial x} \end{align}

Using the chain rule, \begin{align} \frac{\partial}{\partial \theta} \left(\frac{\partial f}{\partial x}\right) &= \frac{\partial x}{\partial \theta}\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial x}\right) \\ &= \frac{\partial x}{\partial \theta} \frac{\partial^2 f}{\partial x^2} \end{align}

and so

$$ \frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\cos\theta\right) = \frac{\cos^2\theta}{r}\frac{\partial^2 f}{\partial x^2} - \sin \theta\frac{\partial f}{\partial x}. $$

Can you take it from here?

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