Answer
a) slope is $\frac{dy}{dx}$ = $\frac{d\sin\theta}{r-d\cos\theta}$
b) prove that $r-d\cos\theta>0$
Work Step by Step
$x$ = $r\theta-d\sin\theta$
$y$ = $r-d\cos\theta$
a)
$\frac{dx}{d\theta}$ = $r-d\cos\theta$
$\frac{dy}{dθ}$ = $d\sin\theta$
$\frac{dy}{dx}$ = $\frac{d\sin\theta}{r-d\cos\theta}$
b)
If $0$ $\lt$ $d$ $\lt$ $r$ then $|d\cos\theta|$ $\leq$ $d$ $\lt$ $r$
so ${r-d\cos\theta}$ $\geq$ $r-d$ $\gt$ $0$
This shows that $\frac{dx}{dθ}$ never vanishes, so the trochoid can have no vertical tangents if $d$ $\lt$ $r$.
