## Calculus: Early Transcendentals 8th Edition

$$\frac{dy}{dx}=\frac{e^\sqrt x}{2\sqrt x}$$
$$y=e^{\sqrt x}$$ $$\frac{dy}{dx}=\frac{d(e^{\sqrt x})}{dx}$$ Let $u=\sqrt x$ and $y=e^u$. Then, according to Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{d(e^u)}{du}\frac{d(\sqrt x)}{dx}$$ $$\frac{dy}{dx}=e^u\times\frac{d(x^{1/2})}{dx}$$ $$\frac{dy}{dx}=e^u\times\frac{1}{2}x^{-1/2}$$ $$\frac{dy}{dx}=e^{\sqrt x}\times\frac{1}{2\sqrt x}$$ $$\frac{dy}{dx}=\frac{e^\sqrt x}{2\sqrt x}$$