Multivariable chain rule exercise

Question

$F:\mathbb{R}^2\rightarrow\mathbb{R}$ 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.

Answer 1

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?