find the derivative of an integral Find $f'(x)$ where $f(x)$ is the integral from ${\sqrt{x}}$ to $x$ of $e^x-e^{t^2} dt$  

Is there an easy way to do this using the fundamental theorem of calculus because if I try to ingretate w.r.t $t$ then $e^{t^2}$ is a bit of a problem.
Apply differentiation under the integral sign, which is a formula derived from FTC:
\begin{align*}
\frac{d}{dx}\int_{\sqrt x}^x(e^x-e^{t^2})dt&amp;=(e^x-e^{x^2})-(e^x-e^x)(\sqrt x)^\prime+\int_{\sqrt x}^x e^x dt \\&amp;= (e^x-e^{x^2})+e^x(x-\sqrt x)
\end{align*}
Yes there is.  Assuming no pathological situations, the derivative of a definite integral is the difference of the integrand evaluated at the two endpoints.
So $$\frac{d}{dx} \int_\sqrt{x}^x \left( e^x - e^{t^2} \right) dt = 
\left. 
e^x - \right) dt = 
\left. \left( e^x - e^{t^2} \right)
\right|_{t=\sqrt{x}}^{t=x} = e^x - e^{x^2}
$$