Derivatives of Inverse Trig Functions — Question 3

Question 3

Let $$f(x)=\arcsin\!\left(\frac{2x}{1+x^2}\right),\qquad x>0.$$

• Express $f$ in terms of $\arctan x$, separately for $0<x\leq1$ and $x>1$.

• Find $f'(x)$ wherever it exists, and analyze $x=1$.

Question 3 - Solution

Put $t=\arctan x$. For $x>0$, $0<t<\pi/2$ and

$$\frac{2x}{1+x^2}=\sin(2t).$$

The principal range of $\arcsin$ is $[-\pi/2,\pi/2]$. Therefore

$$\boxed{f(x)=\begin{cases}2\arctan x,&0<x\le1,\\\pi-2\arctan x,&x>1.\end{cases}}$$

Differentiating on the two open intervals gives

$$\boxed{f'(x)=\begin{cases}2/(1+x^2),&0<x<1,\\-2/(1+x^2),&x>1.\end{cases}}$$

At $x=1$ the left and right derivatives are $1$ and $-1$, respectively, so no derivative exists there.