Differentiation of $\sin^{-1}(1/\sqrt{1+x})$ $$\arcsin \frac{1}{\sqrt{1+x}}.$$ I tried differentiating it but every time my answer came wrong. The answer in my book is $$- \frac{1}{2(x+1)\sqrt x}.$$
Rules I will use:


The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$
The derivative of $x^\alpha$ is $\alpha\cdot x^{\alpha-1}$
The chain rule, i.e. the derivative of $f(g(x))$ is $f'(g(x))\cdot g'(x)$.




First, take $f(x)=\arcsin(x)$ and $g(x)=\frac{1}{\sqrt{1+x}}$, and use the chain rule. The derivative of $f(g(x))$ is

$$\begin{align}f'(g(x))\cdot g'(x)&amp;= \frac{1}{\sqrt{1-g(x)^2}}\cdot g'(x)\\ &amp;= \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+x}}\right)^2}}\cdot g'(x) \\&amp;= \frac{1}{\sqrt{1-\frac{1}{1+x}}}\cdot g'(x)\\&amp;=\frac{1}{\sqrt{\frac{1+x-1}{1+x}}}\cdot g'(x)\\&amp;=\frac{\sqrt{1+x}}{\sqrt x}\cdot g'(x)\end{align}$$

I leave the computation of $g'(x)$ to you.
The function
$$
f(x) = \arcsin{\frac{1}{\sqrt{1+x}}}
$$
can be seen as a composite function $f(x) = u(s(x))$, where $u(x) = \arcsin{x}$ and $s(x)= \frac{1}{\sqrt{1+x}}$. Their derivatives are (I will assume these as given - I won't prove these)
$$
u'(x) = \frac{1}{\sqrt{1-x^2}}
$$
$$
s'(x) =  -\frac{1}{2(x+1)^{2/3}}
$$
Now, the derivative of a composite function is
$$
f'(x) = s'(x) u'(s(x)) = -\frac{1}{2(x+1)^{2/3}} \frac{1}{\sqrt{1-s^2(x)}}
$$
The latter terms is a bit tricky to evaluate, we should take it aside:
$$
\frac{1}{\sqrt{1-s^2(x)}} = \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+x}} \right)^2}}
= \frac{1}{\sqrt{1-\left(\frac{1}{1+x} \right)}}
= \frac{1}{\sqrt{\frac{1+x}{1+x}-\frac{1}{1+x}}}
= \frac{1}{\sqrt{\frac{x}{1+x}}}
= \frac{(1+x)^{1/2}  }{  \sqrt{x} }
$$
therefore, the derivative can be simplified to 
$$
f'(x) = -\frac{1}{2(x+1)^{2/3}} \frac{(1+x)^{1/2}  }{  \sqrt{x} }
 = -\frac{1}{2\sqrt{x}(x+1)}
$$