Partial derivative of nabla operation

Question

Given that $$\nabla^2f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} = \frac{\partial^2f}{\partial x'^2} + \frac{\partial^2f}{\partial y'^2}$$

$$ x = x' \cos \theta - y'\sin \theta \\ y = x' \sin \theta + y'\cos \theta $$

$$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial x'} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x'} $$

I obtain and understand how to get the following $$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta $$

But I don't understand how the following is obtained? i.e. how the $\sinθ\cosθ$ terms.

$$ \frac{\partial^2 f}{\partial x'^2} = \frac{\partial^2 f}{\partial x^2}\cos^2\theta+ \frac{\partial }{\partial x}(\frac{\partial f}{\partial y})\sin\theta\cos\theta + \frac{\partial }{\partial y}(\frac{\partial f}{\partial x})\cos\theta\sin\theta + \frac{\partial^2 f}{\partial y^2}\sin^2\theta $$

Answer 1

You have shown that $$ \frac{\partial f}{\partial x'} = \cos\theta \frac{\partial f}{\partial x} + \sin\theta \frac{\partial f}{\partial y}. $$ Note that you can rewrite this as $$ \frac{\partial}{\partial x'} = \left(\cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y}\right) f. $$ Therefore you have found the derivative w.r.t. $x'$ in terms of the derivatives w.r.t. $x$ and $y$, namely $$ \frac{\partial}{\partial x'} = \cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y}. $$ Therefore $$ \begin{align*} \frac{\partial^2 f}{\partial x'^2} &=\frac{\partial}{\partial x'} \frac{\partial f}{\partial x'} \\ &= \left(\cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y}\right)\left( \cos\theta \frac{\partial f}{\partial x} + \sin\theta \frac{\partial f}{\partial y}\right) \\ &= \cos^2\theta \frac{\partial}{\partial x}\frac{\partial f}{\partial x} + \sin \theta\cos\theta \frac{\partial}{\partial y}\frac{\partial f}{\partial x} + \cos \theta \sin\theta \frac{\partial}{\partial x}\frac{\partial f}{\partial y} + \sin^2\theta \frac{\partial }{\partial y}\frac{\partial f}{\partial y}. \end{align*} $$

Answer 2

All you need to do is to consider $\frac{\partial f}{\partial x'}=F$, and then apply the following formula $$ \frac{\partial F}{\partial x'} = \frac{\partial F}{\partial x} \cos \theta + \frac{\partial F}{\partial y} \sin \theta $$

Answer 3

Start from $$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta $$ You replace $f$ everywhere with $ \frac{\partial f}{\partial x'}$ $$ \frac{\partial }{\partial x'}\frac{\partial f}{\partial x'} = \frac{\partial }{\partial x} \frac{\partial f}{\partial x'}\cos \theta + \frac{\partial }{\partial y}\frac{\partial f}{\partial x'} \sin \theta $$ Now plug in $\frac{\partial f}{\partial x'}$ on the right side