Answer
$\frac{u}{\sqrt {1-u^2}}$
Work Step by Step
Step 1. Letting $sin^{-1}u=t$, we have $sin(t)=u$ and $-\frac{\pi}{2}\le t \le \frac{\pi}{2}$
Step 2. Form a right triangle with sides $y=|u|,r=1,x=\sqrt {1-u^2}$ and angle $|t|$ facing $y$.
Step 3. As $tan(t)$ and $sin(t)$ have the same sign, we have $tan(sin^{-1}u)=tan(t)=\frac{y}{x}=\frac{u}{\sqrt {1-u^2}}$
