Differentiate $f(x)=\int_{\sqrt{x}}^{\sin^2 x}\arctan (1+e^{t^2})dt$

We have 
\[f'(x)=(\sin^2 x)' \arctan (1+e^{(\sin^2x)^2})-(x^{1/2})'\arctan (1+e^{(\sqrt{x})^2})\]
\[=2\sin x \cos x \arctan (1+e^{\sin^4x})-\frac{1}{2}x^{-1/2}\arctan (1+e^{x})\]
\[=2\sin x \cos x \arctan (1+e^{\sin^4x})-\frac{1}{2 \sqrt{x}}\arctan (1+e^{x}).\]