Math, asked by balaganeshj2004, 7 months ago

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

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Answers

Answered by tannumishra1464
0

Step-by-step explanation:

this might be the correct answer

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Answered by senboni123456
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} )}

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