Question

$Ify=co{s}^{-1}\frac{1-x^2}{1+x^2}then\frac{dy}{dx}atx=-2willbe$

• A. $\frac{2}{5}$
• B. $-\frac{2}{5}$
• C. Does not exist
• D. None of the above

Answer 1

The correct option is B $-\frac{2}{5}$

Let’s first do it by putting $x=tan\theta or\theta =ta{n}^{-1}x$ Doing this we get $y=co{s}^{-1}\frac{(1-ta{n}^2\theta )}{1+ta{n}^2\theta }$ .
But, $\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=cos2\theta$ (It’s a standard trigonometric identity).
$i.e.y=co{s}^{-1}cos2\theta Nowco{s}^{-1}cosx=xi.e.co{s}^{-1}cos2\theta =2\theta So\frac{dy}{dx}=\frac{2}{1+x^2}\left[Using\frac{d}{dx}\right(ta{n}^{-1}x)=\frac{1}{1+x^2}]$
The solution looks correct but it’s not.
Look $\theta =ta{n}^{-1}x$ implies that $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$
Also $co{s}^{-1}cosx=x$ only when $0\le x\le \pi$
But observe that in the above solution we have used $co{s}^{-1}cosx=x$ for the entire domain which is not correct. Let’s correct it then.
$co{s}^{-1}cos2\theta =2\theta onlywhen\theta \le 2\theta \le \pi or0\le \theta \le \frac{\pi }{2}$
So for $0\le \theta \le \frac{\pi }{2}$, the domain of $ta{n}^{-1}x$ i.e. x will be 0 to $\infty$
$i.efor0\le \theta \le \frac{\pi }{2},0\le x<\infty$
So the above result is correct only for positive x and is not correct for negative x.
Also $co{s}^{-1}cos2\theta =-2\theta if-\pi \le 2\theta <0or-\frac{\pi }{2}\le \theta <0i.e.,-\infty <x<0$ (Check the domain of $ta{n}^{-1}xfor-\frac{\pi }{2}\le \theta <0$)
So for negative x
$co{s}^{-1}cos2\theta =-2\theta =-2ta{n}^{-1}xi.e.forx<0y=-2ta{n}^{-1}xSo\frac{dy}{dx}=\frac{-2}{1+x^2}forx<0\frac{2}{1+x^2}forx\ge 0$
To get its value at -2, we will put x = -2 in $\frac{-2}{1+x^2}$ because this is the derivate for x < 0.
${So\frac{dy}{dx}|}_{x=-2}=\frac{2}{1+4}=\frac{-2}{5}$
Using chain Rule.
$y=co{s}^{-1}\frac{1-x^2}{1+x^2}\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(\frac{1-x^2}{1+x^2}\right)}^2}}\times \frac{-2x(1+x^2)-(1-x^2)2x}{(1+x^2{)}^2}$
[Using chain rule and standard formula for derivative of $co{s}^{-1}x$]
$=\frac{(1+x^2)}{\sqrt{(1+x^2{)}^2}-(1+x^2{)}^2}\times \frac{2x.2}{(1+x^2{)}^2}=\frac{4x}{\sqrt{4x^2}(1+x^2)}=\frac{2x}{|x|(1+x^2)}=\frac{4x}{\sqrt{4x^2}(1+x^2)}=\frac{2x}{|x|(1+x^2)}$ (Please note that $\sqrt{4x^2}=2|x|andnot2x)$
So $\frac{dy}{dx}=\left\{\begin{array}{cc}\frac{-2}{1+x^2}forx<0& \\ doesnotexistatx=0& \\ \frac{2}{1+x^2}whenx>0& \end{array}\right\}$
$As|x|=xforx>0and-xforx<0$