Multivariable partial differentiation 
$f(r,\theta)=r^5\cos \theta$,$x=r\cos \theta$, $y=r\sin \theta$
Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial x}$ in therms of $r$ and $\theta$

I know that $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial r}\cdot \frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\cdot \frac{\partial \theta}{\partial x}$$
so I have $\frac{\partial f}{\partial r}=5r^4\cos \theta$
$x=r\cos \theta \Rightarrow r=\frac{x}{cos\theta}$
so $\frac{\partial r}{\partial x}=\frac{1}{\cos \theta}$
Or because $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}(\frac{y}{x})$
It is $r=\frac{x}{\cos \theta}=\frac{x}{\frac{x}{\sqrt{x^2+y^2}}}=\sqrt{x^2+y^2}$ so $\frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=cos\theta$
?

\begin{align}
f &= r^5 \cos \theta = r^4 \cdot r \cos \theta = x \cdot \left ( x^2 + y^2\right)^2 \implies \\
f_x &= \left ( x^2 + y^2\right)^2 + x \cdot 2(x^2 + y^2) \cdot 2x = \left ( x^2 + y^2\right) \left( x^2 + y^2 + 4x^2\right) = \\
&= \left ( x^2 + y^2\right) \left ( 5x^2 + y^2\right) = r^4 \left ( 5 \cos^2 \theta + \sin^2 \theta\right) \\
f_y &= 4x y \left ( x^2 + y^2\right) = 4r^2 \cos \theta \sin \theta \cdot r^2 = 4r^4 \cos \theta \sin \theta
\end{align}
You can simplify even more, if you want, using trig identities.
