Differentiation of $\sin^{-1}(1/\sqrt{1+x})$

Question

$$\arcsin \frac{1}{\sqrt{1+x}}.$$ I tried differentiating it but every time my answer came wrong. The answer in my book is $$- \frac{1}{2(x+1)\sqrt x}.$$

Answer 1

Rules I will use:

1. The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$
2. The derivative of $x^\alpha$ is $\alpha\cdot x^{\alpha-1}$
3. 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

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

I leave the computation of $g'(x)$ to you.

Answer 2

If you set $y$ equal to your expression and take $\sin$ of both side you have

$$\sin y = \frac{1}{\sqrt{1+x}}.$$

Do implicit differentiation to get

$$(\cos y) y'= \frac{-1}{2(1+x)^{3/2}}.$$

Using the first equation we get

$$\cos y = \sqrt{1-\sin^2 y }= \sqrt{1- \left(\frac{1}{\sqrt{1+x}}\right)^2 }= \sqrt{\frac{x}{1+x}}.$$

Plug this in the first equation and solve for $y'.$

Answer 3

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 $$ f'(x) = -\frac{1}{2(x+1)^{2/3}} \frac{(1+x)^{1/2} }{ \sqrt{x} } = -\frac{1}{2\sqrt{x}(x+1)} $$