Question

Differentiate ${\cot }^{-1}\frac{2\sqrt{1+x^2}-5\sqrt{1-x^2}}{5\sqrt{1+x^2}+2\sqrt{1-x^2}}$ with respect to ${\cos }^{-1}\sqrt{(1-x^4)}.$

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $-\frac{1}{2}$

Let $y={\cot }^{-1}\frac{2\sqrt{1+x^2}-5\sqrt{1-x^2}}{5\sqrt{1+x^2}+2\sqrt{1-x^2}}={\tan }^{-1}\frac{5\sqrt{1+x^2}+2\sqrt{1-x^2}}{2\sqrt{1+x^2}-5\sqrt{1-x^2}}$

Divide numerator and denominator by $2\sqrt{1+x^2}$

$\therefore y={\tan }^{-1}\frac{\frac{5}{2}+\sqrt{\frac{1-x^2}{1+x^2}}}{1-\frac{5}{2}\sqrt{\frac{1-x^2}{1+x^2}}}$

or $y={\tan }^{-1}\frac{5}{2}+{\tan }^{-1}\sqrt{\frac{1-x^2}{1+x^2}}$

and Let $z={\cos }^{-1}\sqrt{1-x^4}$

Put $x^2=\cos 2\theta$

$\therefore y=c+{\tan }^{-1}\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}$ , where $c={\tan }^{-1}\frac{5}{2}$

$=c+{\tan }^{-1}\left(\tan \theta \right)=c+\theta$

Now $z={\cos }^{-1}\left(\sin 2\theta \right)=\frac{\pi }{2}-2\theta =\frac{\pi }{2}-2(y-c).$ or $z=\frac{\pi }{2}+2c-2y$

We have to find the value of $\frac{dy}{dz}.$
$\therefore \frac{dy}{dz}=-\frac{1}{2}$