Find $\frac{du}{dv}$ if $u=\sin^{-1}\frac{2x}{1+x^2}$ and $v=\tan^{-1}\frac{2x}{1-x^2}$

Question

If $u=\sin^{-1}\frac{2x}{1+x^2}$ and $v=\tan^{-1}\frac{2x}{1-x^2}$, then find $\frac{du}{dv}$

My Attempt $$ \frac{dv}{dx}=\frac{2}{1+x^2} $$ $$ \frac{du}{dx}=\frac{1}{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}.\frac{(1+x^2).2-2x.2x}{(1+x^2)^2}=\frac{|1+x^2|}{\sqrt{1+x^4+2x^2-4x^2}}.\frac{2+2x^2-4x^2}{(1+x^2)^2}\\ =\frac{|1+x^2|}{|1-x^2|}.\frac{2(1-x^2)}{(1+x^2)^2}=\frac{2(1-x^2)}{|1-x^2|.(1+x^2)}=\begin{cases}\frac{2}{1+x^2},\;x^2\leq1\\\frac{-2}{1+x^2},\;x^2>1\\\end{cases} $$ Mathematica is giving after Simplify and PoweExpand, $\frac{du}{dx}=\frac{-2}{1+x^2}$

But I think $\frac{du}{dx}=\frac{2}{1+x^2}$ as $\sin^{-1}$ is an increasing function and thus $\frac{du}{dv}=1$. How do I find the required derivative ?

Answer 1

If $x= \tan u$. Then $\displaystyle \sin2u=\frac{2x}{1+x^2}$ and $\displaystyle \tan2u=\frac{2x}{1-x^2}$.

Let $\displaystyle y=\arctan x$.

If $|x|<1$, then $\displaystyle |y|<\frac{\pi}{4}$ and hence $\displaystyle |2y|<\frac{\pi}{2}$. $u=\arcsin(\sin 2y)$ and $v=\arctan(\tan 2y)$ imply that $u=2y=v$.

If $x>1$, then $\displaystyle \frac{\pi}{4}<y<\frac{\pi}{2}$ and hence $\displaystyle \frac{\pi}{2}<2y<\pi$. Since $u=\arcsin(\sin 2y)$, $\displaystyle 0<u<\frac{\pi}{2}$ and $\sin u=\sin 2y=\sin(\pi-2y)$. So, $u=\pi-2y$. Since $v=\arctan(\tan 2y)$, $\displaystyle \frac{-\pi}{2}<v<0$ and $\tan v=\tan 2y=\tan(2y-\pi)$. So, $v=2y-\pi$.

If $x<-1$, then $\displaystyle \frac{-\pi}{2}<y<\frac{-\pi}{4}$ and hence $\displaystyle -\pi<2y<\frac{-\pi}{2}$. Since $u=\arcsin(\sin 2y)$, $\displaystyle \frac{-\pi}{2}<u<0$ and $\sin u=\sin 2y=\sin(-\pi-2y)$. So, $u=-\pi-2y$. Since $v=\arctan(\tan 2y)$, $\displaystyle 0<v<\frac{\pi}{2}$ and $\tan v=\tan 2y=\tan(2y+\pi)$. So, $v=2y+\pi$.

We have that $u=v$ if $|x|<1 $ and $u=-v$ if $|x|>1 $.

$$ \frac{du}{dv}=\begin{cases} 1 &\textrm{if }|x|<1 \\ -1 &\textrm{if }|x|>1\end{cases}$$

The graph of $\displaystyle u=\arcsin\frac{2x}{1+x^2}$:

The graph of $\displaystyle v=\arctan\frac{2x}{1-x^2}$:

Answer 2

Using my answer in Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$2\arctan x=\begin{cases} \arctan\frac{2x}{1-x^2} &\mbox{if } x^2<1\\ \pi+\arctan\frac{2x}{1-x^2} & \mbox{if } x^2>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x^2=1\end{cases} $$

So for $x\ne1,$ $$\dfrac{d\left(\arctan\frac{2x}{1-x^2}\right)}{d(\arctan x)}=2\ \ \ \ (1)$$

Similarly,

putting $\arctan x= y,x=\tan y$

$$\arcsin\dfrac{2x}{1+x^2}=\arcsin(\sin2y)=\begin{cases}2y &\mbox{if } -\dfrac\pi2\le2y\le\dfrac\pi2\iff x^2\le1\\ \pi-2y & \mbox{if } \dfrac\pi2\le2y\le\pi\iff x>1\\-\pi-2y & \mbox{if } -\pi\le 2y\le-\dfrac\pi2\iff x<-1\end{cases}$$

$$\implies\dfrac{d\left(\arcsin\frac{2x}{1+x^2}\right)}{d(\arctan x)}=2\iff x^2\le1\ \ \ \ (2)$$

Divide $(1)$ by $(2)$

Answer 3

To answer your question, $u$ increases for $|x|<1$ and decreases for $|x|>1$. Note: $$\sin u=\frac{2x}{1+x^2}=g(x)$$ $$g'(x)=0 \Rightarrow g(-1)=-1 \ (\text{global min}); \ g(1)=1 \ (\text{global max}).$$ $$g'(x)>0, \ |x|<1; \ \ g'(x)<0, \ |x|>1.$$