1. Problem: Find the first derivative $\frac{dy}{dx}$ for $y=2^x \sin^{-1} x$. Use the product rule since $y$ is a product of two functions. 2. Formula: Product rule states $\frac{d}{dx}[u v] = u' v + u v'$. Here, $u=2^x$, $v=\sin^{-1} x$. 3. Derivatives: $u' = 2^x \ln 2$, $v' = \frac{1}{\sqrt{1-x^2}}$. 4. Apply product rule: $$\frac{dy}{dx} = 2^x \ln 2 \cdot \sin^{-1} x + 2^x \cdot \frac{1}{\sqrt{1-x^2}}$$ 5. Final answer: $$\boxed{\frac{dy}{dx} = 2^x \ln 2 \sin^{-1} x + \frac{2^x}{\sqrt{1-x^2}}}$$ This derivative combines the exponential growth of $2^x$ and the rate of change of the inverse sine function. The product rule allows us to differentiate each part separately and sum their contributions.