My attempt:
$\pi=180^{\circ}$
$x^{\circ}=\frac{\pi x}{180}$
$$\frac{d}{dx}(\ln\cos x^{\circ})$$
$$=\frac{d}{dx}(\ln\cos \frac{\pi x}{180})$$
$$=-\frac{1}{\cos \frac{\pi x}{180}}\cdot(\sin\frac{\pi x}{180})\frac{\pi}{180}$$
$$=-\frac{\pi}{180}\tan x^{\circ}$$
But my book says the answer is $-\tan x^{\circ}$. Isn't my book wrong?
Your book probably wants you to differentiate $\ln \cos x$ with respect to $x$, since $$ \frac{\mathrm d}{\mathrm d x} (\ln \cos x) = - \tan x$$ Moreover, it is a very outlandish differentiation problem which further increases the possibility that it is a typo (either in the answer or in the question itself). Otherwise, your answer is correct, and $$ \frac{\mathrm d}{\mathrm d x} \left(\ln \cos \frac{\pi x}{180} \right) = - \frac{\pi}{180} \, \tan \frac{\pi x}{180}$$
