Differentiation answer check $$f(x)=\arcsin \sqrt{\frac{x}{x+1}} + \arctan \sqrt{x} \mbox{.}$$
$$f'(x) = \frac{1}{\sqrt{1-\frac{x}{x+1}}} \cdot \frac{1}{2}\sqrt{\frac{x+1}{x}} \cdot -x^{-2} + \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} = -\frac{\sqrt{x+1}}{2\sqrt{1-\frac{x}{x+1}}{\sqrt{x}}\frac{1}{x^2}} + \frac{1}{2\sqrt{x} + 2x}\mbox{.}$$
Is there anything else I can do with the answer?

HINT:
Let $\displaystyle\arcsin\sqrt{\frac x{x+1}}=y$
$$\implies \sqrt{\frac x{x+1}}=\sin y\implies \frac x{x+1}=\sin^2y$$
$$\implies x=\tan^2y\implies \sqrt x=\tan y \implies y=\arctan \sqrt x$$

Alternatively, let $\displaystyle\arctan \sqrt x=u\implies \sqrt x=\tan u\implies x=\tan^2u$
$\displaystyle\implies \frac x{x+1}=\frac{\tan^2u}{1+\tan^2u}=\frac{\tan^2u}{\sec^2u}=\sin^2u$
$\displaystyle\implies \arcsin\sqrt{\frac x{x+1}}=\arcsin (\sin u)=u=\arctan \sqrt x$
