Question

can u guys help me out of this question plz I need answer for this help
i will mark branliest ​

Answer 1

Step-by-step explanation:

this might be the correct answer

Answer 2

Step-by-step explanation:

We have,

$y = \tan^{ - 1} ( \frac{ \sqrt{1 + {x}^{2} } + 1}{x} )$

Let, x = tan(θ)

=>θ = tan^(-1) x

$= > y = \tan^{ - 1} ( \frac{ \sqrt{1 + \tan^{2} ( \theta) } + 1 }{ \tan( \theta) } )$

$= > y = \tan^{ - 1} ( \frac{ \sec( \theta) - 1 }{ \tan( \theta) } )$

$= > y = \tan^{ - 1} ( \frac{1 - \cos( \theta) }{ \sin( \theta) } )$

$= > y = \tan^{ - 1} ( \frac{2 \sin ^{2} ( \frac{ \theta}{2} ) }{2 \sin( \frac{ \theta}{2} ) \cos( \frac{ \theta}{2} ) } )$

$= > y = \tan^{ - 1} ( \tan( \frac{ \theta}{2} ) )$

$= > y = \frac{ \theta}{2}$

$= > y = \frac{1}{2} \tan^{ - 1} (x)$

Now,

$\frac{dy}{dx} = \frac{1}{2} \times \frac{1}{1 + {x}^{2} }$

$= > \frac{dy}{dx} = \frac{1}{2(1 + {x}^{2} )}$