On differentiating tan^(−1) [ (√(1+x^2) −1)/x ] with respect to x, the result would be

Correct answer: (A) 1/2 .1/(1+x^2). Let us assume that, y=tan^(−1) [ (√(1+x^2) −1)/x ] Put : x=tanθ ∴ (√(1+x^2)−1)/x=(√(1+tan^2θ)−1)/(tanθ)=(secθ−1)/(tanθ) =(1−cosθ)/(sinθ)=(2sin^2θ2)/(2sinθ2cosθ2) =tan(θ)/2 On differentiating with respect to x: ∴ y=tan^(−1) ( tan(θ)/2 )=1/2 θ=1/2 tan^(−1) x ⇒ dy/dx =1/2 .1/(1+x^2) .

Q39 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

On differentiating $\tan^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]$ with respect to $x$ , the result would be

Correct answer
A
$\frac{1}{2} . \frac{1}{1 + x^{2}}$
B
$\frac{1}{1 + x^{2}}$
C
$\frac{2}{1 + x^{2}}$
D
$\frac{1}{2} . \frac{1}{1 + 2 x}$

Solution

Let us assume that,

$y = \tan^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]$

Put : $x = \tan θ$

$\begin{array}{l} ∴ \frac{\sqrt{1 + x^{2}} - 1}{x} = \frac{\sqrt{1 + \tan^{2} θ} - 1}{\tan θ} = \frac{\sec θ - 1}{\tan θ} \\ = \frac{1 - \cos θ}{\sin θ} = \frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}} \\ = \tan \frac{θ}{2} \end{array}$

On differentiating with respect to x:

$\begin{array}{l} ∴ y = \tan^{- 1} (\tan \frac{θ}{2}) = \frac{1}{2} θ = \frac{1}{2} \tan^{- 1} x \\ \Rightarrow \frac{d y}{d x} = \frac{1}{2} . \frac{1}{1 + x^{2}} . \end{array}$

Hence, the correct answer is option: 'A'.

On differentiating tan −1 [ 1+ x 2 −1 x ] with respect to x, the result would be

Solution

More Continuity and Differentiability Questions

On differentiating $\tan^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]$ with respect to $x$ , the result would be