$u = f(x, y).$ $u$ is a C1 function. $x = r\cos\theta$ and $y = r\sin\theta$. Show that
$$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\cos\theta \pm \frac{\partial u}{\partial \theta} \frac{\sin \theta}{r} $$
Attempt:
I thought I'd calculate $\frac{\partial u}{\partial r}$ then isolate $\frac{\partial u}{\partial x}$ in the following manner
$$ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \tag{1} $$
$$ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta \tag{2} $$
Using $(2)$, when you isolate the requested derivative, however, you get
$$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\frac{1}{\cos \theta} - \frac{\partial u}{\partial y} \tan\theta \tag{3} $$
Other attempt:
Obviously you get $\frac{\partial u}{\partial x}$ in the other partial derivative $\frac{\partial u}{\partial \theta}$. Then I isolated $ \frac{\partial u}{\partial x}$ there too, but it's not the requested result.
Also if you add the $\frac{\partial u}{\partial x}$ from the two partial derivatives then divide by 2, it's still not the requested result. So.. I have no idea what kind of manipulation would give the requested result.
you have full-derivative
$$\mathrm{d}u = \frac{\partial u}{\partial r}\mathrm{d}r + \frac{\partial u}{\partial \theta}\mathrm{d}\theta \tag1$$
$$\mathrm{d}u = \frac{\partial u}{\partial x}\mathrm{d}x + \frac{\partial u}{\partial y}\mathrm{d}y \tag2$$
then you reverse the relation between $(x,y)$ and $(r,\theta)$
$$r^2=x^2+y^2 \quad \text{and} \quad \tan\theta=\frac{y}{x}$$
and put this into $\mathrm{d}u$ in (1)
$$\mathrm{d}r = \frac{\partial r}{\partial x}\mathrm{d}x + \frac{\partial r}{\partial y}\mathrm{d}y$$
$$\mathrm{d}\theta = \frac{\partial\theta}{\partial x}\mathrm{d}x + \frac{\partial\theta}{\partial y}\mathrm{d}y$$
thus
$$\mathrm{d}u = \left(\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial\theta}\frac{\partial\theta}{\partial x}\right)\mathrm{d}x + \left(\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial\theta}\frac{\partial\theta}{\partial y}\right)\mathrm{d}y$$
in the same position of $\mathrm{d}x$ in (2) which is
$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial\theta}\frac{\partial\theta}{\partial x}$$
that is answer you want.