<s>[INST] Polar equation representation of an exponential spiral involving $ \arctan(z) $ An exponential spiral is formulated as:
$$ \arctan(z) = \ln\left(\sqrt{1+z^2}\right)+\ln(x)+C $$ with: $$z=y/x$$
Represent the equation in polar coordinates and show $y' = \tan(\pi/4 + \theta)$.
Substituting in $ x = r\cos(\theta) $ and $ y = r\sin(\theta) $ I get:
$$ \arctan(z) = \ln\left(\sqrt{1+\frac{r^2\sin^2(\theta)}{r^2\cos^2(\theta)}}\,\right) + \ln(r\cos(\theta)) + C $$
$$ \arctan(z) = \ln\left(\sqrt{1+\tan^2(\theta)}\,\right) + \ln(r\cos(\theta)) + C $$
I'm not sure how to do this problem. I get stuck here.
 [/INST] If $x = r\cos\theta, y = r\sin\theta$, assuming we are in the first/fourth quadrant $\cos\theta >0$, otherwise your equation $\ln x$ would not make sense. Then the polar coordinates can be written as:
$$
r = \sqrt{x^2 + y^2}, \text{ and } \theta = \arctan\frac{y}{x}\tag{1}
$$
You have arrived at 
$$ \arctan\frac{y}{x} = \ln\sqrt{1+\tan^2\theta}\, + \ln(r\cos\theta) + C \tag{2}$$
Plugging (1) into (2), exploiting the trigonometry identity $1+\tan^2\theta = \sec^2\theta$:
$$
\theta = \ln\sec\theta + \ln(r\cos\theta) + C
$$
which is
$$
\theta = \ln(\sec\theta\cdot r\cos\theta) +C = \ln r +C
$$
hence:
$$
r = c e^\theta \tag{3}
$$
and this is the equation for the exponential spiral. Plugging (3) back into the original polar coordinate parametrization:
$$
x = ce^\theta \cos\theta, \text{ and } y = c e^\theta\sin\theta 
$$
For the $y'$:
$$
y' = \frac{dy}{dx} =  \frac{dy/d\theta}{dx/d\theta} = \frac{\sin\theta + \cos\theta}{\cos\theta - \sin\theta} 
$$
To get a tangent out, simply exploiting the formula:
$$\begin{aligned}
\cos(\theta +\frac{\pi}{4}) = \cos\theta \cos\frac{\pi}{4}- \sin\theta\sin\frac{\pi}{4}
\\
\sin(\theta +\frac{\pi}{4}) = \sin\theta \cos\frac{\pi}{4} + \cos\theta\sin\frac{\pi}{4}
\end{aligned}
$$
Therefore:
$$
y' = \frac{\sin(\theta +\frac{\pi}{4}) }{\cos(\theta +\frac{\pi}{4}) } = \tan(\theta +\frac{\pi}{4})
$$
</s>