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# If $f(x)={{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right)$ , then(a) $f$ is derivable for all $x$, with $\left| x \right|<1$(b) $f$ is not derivable at $x=1$(c) $f$ is not derivable at $x=-1$(d) $f$ is derivable for all $x$, with $\left| x \right|>1$

Last updated date: 15th Jul 2024
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Hint: Check the differentiability of f(x) at the end points of its domain and check which option is matching with your answer. Also use the half angle formula in terms of “tan” for substitution.

In a given problem we have to find whether the function is differentiable and if yes then at what values?
For that we will just rewrite given equation,
$f(x)={{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right)$
Now, to simplify the problem substitute
$x=\tan \theta$ In the above problem. Therefore, $\theta ={{\tan }^{-1}}x$…………………………………. (1)
$\therefore f(x)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+\mathop{\tan \theta }^{2}} \right)$
To proceed further we should know the Half Angle formula for $\sin 2\theta$ which is given below,
Formula:
$\sin 2\theta =\dfrac{2\tan \theta }{1+\mathop{\tan \theta }^{2}}$
Therefore $f(x)$ will become,
$\therefore f(x)={{\sin }^{-1}}\left( \sin 2\theta \right)$ ……………………….. (2)
If we have to simplify further then we should know it’s simplification in various domains, which are given below,
Formulae:
${{\sin }^{-1}}\left( \sin x \right)=-\pi -x$ For $x<\dfrac{-\pi }{2}$
For $\dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2}$
${{\sin }^{-1}}\left( \sin x \right)=\pi -x$ For $x>\dfrac{\pi }{2}$
We can write equation (2) according to above formulae by replacing ‘x’ with $2\theta$
As, $f(x)={{\sin }^{-1}}\left( \sin 2\theta \right)$
$f(x)=-\pi -2\theta$ For $2\theta <\dfrac{-\pi }{2}$……………………………. (2)
$f(x)=2\theta$ For $\dfrac{-\pi }{2}\le 2\theta \le \dfrac{\pi }{2}$…………………….. (3)
$f(x)=\pi -2\theta$ For $2\theta >\dfrac{\pi }{2}$……………………………… (4)
Before substituting the value of $\theta$ we will first convert limits,
As, $\dfrac{-\pi }{2}\le 2\theta \le \dfrac{\pi }{2}$
Dividing by 2 we will get,
$\dfrac{-\pi }{4}\le \theta \le \dfrac{\pi }{4}$
Take tangent of all angles,
$\tan \dfrac{-\pi }{4}\le \tan \theta \le \tan \dfrac{\pi }{4}$
$\therefore -\tan \dfrac{\pi }{4}\le \tan \theta \le \tan \dfrac{\pi }{4}$
$\therefore -1\le \tan \theta \le 1$
From (1) we can write above equation as,
$\therefore -1\le x\le 1$………………………………… (5)
Now, we can easily write equations (2), (3), (4) by substituting $x=\tan \theta$ from (1) and replacing limits with the help of (5),
$f(x)=-\pi -2\tan x$ For $x<-1$
$f(x)=2\tan x$ For $-1\le x\le 1$
$f(x)=\pi -2\tan x$ For $x>1$
Now we will check the differentiability at -1, for that we are going to use the formula given below for several times.
Formula:
$\dfrac{d}{dx}\tan x=\dfrac{1}{1+\mathop{x}^{2}}$
$L.H.D.={{\left[ \dfrac{d}{dx}(-\pi -2\tan x) \right]}_{x=-1}}={{\left[ \dfrac{-2}{1+\mathop{x}^{2}} \right]}_{x=-1}}=\dfrac{-2}{2}=-1$$\dfrac{d}{dx}\tan x=\dfrac{1}{1+\mathop{x}^{2}}$
$R.H.D.={{\left[ \dfrac{d}{dx}(2\tan x) \right]}_{x=-1}}={{\left[ \dfrac{2}{1+\mathop{x}^{2}} \right]}_{x=-1}}=\dfrac{2}{2}=1$
$\therefore L.H.D.\ne R.H.D.$
Therefore f(x) is not differentiable at -1…………………………………………. (6)
$L.H.D.={{\left[ \dfrac{d}{dx}(2\tan x) \right]}_{x=1}}={{\left[ \dfrac{2}{1+\mathop{x}^{2}} \right]}_{x=1}}=\dfrac{2}{2}=1$
$L.H.D.={{\left[ \dfrac{d}{dx}(\pi -2\tan x) \right]}_{x=1}}={{\left[ \dfrac{-2}{1+\mathop{x}^{2}} \right]}_{x=1}}=\dfrac{-2}{2}=-1$
$\therefore L.H.D.\ne R.H.D.$
Therefore f(x) is not differentiable at 1………………………………………….. (7)

As, f(x) is not differentiable at $x=1$ and $x=-1$ we can say that f(x) is only differentiable only in its domain with open intervals i.e. In $(-1,1)$.
$\because$[From (6) and (7)]
The domain can also be expressed as $\left| x \right|<1$
This can be shown as follows,
$\left| x \right|<1\equiv$ $x<1$ And $-x<1$
$\equiv$$x\in [0,1)$ And $x>-1$
$\equiv$$x\in [0,1)$ And $x\in (-1,0]$
$\left| x \right|<1$ $\equiv$ $x\in (-1,1)$
Option (a) (b) and (c) are the correct answers.

Note:
Convert the limits very much carefully as there are chances of silly mistakes.
We should know how the functions can be defined in different domains as given below,
$f(x)=-\pi -2{{\tan }^{-1}}x$ For $x<-1$
$f(x)=2{{\tan }^{-1}}x$ For $-1\le x\le 1$
$f(x)=\pi -2{{\tan }^{-1}}x$ For $x>1$