FInd the derivative of $f(x)=\sqrt{x+\arctan x}$

We compute 
\[f'(x)=(x+\arctan x)^{\frac{1}{2}}=\frac{1}{2}(x+\arctan x)'(x+\arctan x)^{\frac{1}{2}-1}\]
\[=\frac{1}{2}(1+\frac{1}{1+x^2})(x+\arctan x)^{-\frac{1}{2}}\]
\[=\frac{1}{2}(1+\frac{1}{1+x^2}) \frac{1}{\sqrt{x+\arctan x}}.\]
Here we have used the derivative formulas 
\[(u^n)'=n u' u^{n-1}\]
and
\[(\arctan x)'=\frac{1}{1+x^2}.\]