## Calculus (3rd Edition)

$$f'(t)= \frac{1}{2} \frac{1}{\sqrt{t}} e^{\sqrt{t}} .$$
Recall that $(e^x)'=e^x$ Since we have $$f(t)= e^{\sqrt{t}}$$ then the derivative $f'(t)$, using the chain rule, is given by $$f'(t)=e^{\sqrt{t}} (\sqrt{t})'=\frac{1}{2} \frac{1}{\sqrt{t}} e^{\sqrt{t}} .$$