Question

If y = \(\rm \cos ^ {-1} \left(\frac {1 - x^2}{1 + x^2}\right)\), then find \(\rm\frac{dy}{dx}\).
1. \(\rm 1\over {1 + x^2}\)
2. \(\rm 2\over {1 + x^2}\)
3. \(\rm 2\over {2 + x^2}\)
4. None of these.

Answer 1

Correct Answer - Option 2 : \(\rm 2\over {1 + x^2}\)

Concept:

Derivatives of Inverse Trigonometric Functions:

\(\begin{matrix}\rm \frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}} & \rm \frac{d}{dx}\left(\csc^{-1}x\right)=\frac{-1}{|x|\sqrt{x^2-1}} \\ \rm \frac{d}{dx}\left(\cos^{-1}x\right)=\frac{-1}{\sqrt{1-x^2}} & \rm \frac{d}{dx}\left(\sec^{-1}x\right)=\frac{1}{|x|\sqrt{x^2-1}} \\ \rm \frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2} & \rm \frac{d}{dx}\left(\cot^{-1}x\right)=\frac{-1}{1+x^2} \end{matrix} \)

Trigonometric Formulae:

\(\rm \sin 2x = \frac{2\tan x}{1+\tan^2x}\)

\(\rm \cos 2x = \frac{1-\tan^2x}{1+\tan^2x}\)

\(\rm \tan 2x = \frac{2\tan x}{1-\tan^2x}\)

 

Calculation:

We have y = \(\rm \cos ^ {-1} \left(\frac {1 - x^2}{1 + x^2}\right)\).

Let x = tan z.

∴ y = \(\rm \cos ^ {-1} \left(\frac {1 - \tan^2z}{1 + \tan^2z}\right)\) = cos^-1 (cos 2z) = 2z = 2 tan^-1 x.

Differentiating w.r.t. x, we get:

\(\rm\frac{dy}{dx}= \frac{d}{dx}\left(2\tan^{-1}x\right)=\frac{2}{1+x^2}\).