Find dy - dx in the following questions-y-sin -12 x-1-x2

Given, y=sin^ 1(2x/1+x^2) putting θ = tan^ 1x i.e., x=tan θ Then, y =sin^ 1( 2tant θ/1+tan^2θ) =sin^ 1(sin 2θ) =2θ =2tant^ 1x [∵ sin 2θ = 2tant θ/ ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

Find dy / dx ...

Question

Find $\frac{d y}{d x}$ in the following questions:

$y = s i n^{- 1} (\frac{2 x}{1 + x^{2}})$

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Solution

Given, $y = s i n^{- 1} (\frac{2 x}{1 + x^{2}})$

putting $θ = t a n^{- 1} x i . e ., x = t a n θ$

Then, $y = s i n^{- 1} (\frac{2 t a n t θ}{1 + t a n^{2} θ}) = s i n^{- 1} (s i n 2 θ) = 2 θ = 2 t a n t^{- 1} x [∵ s i n 2 θ = \frac{2 t a n t θ}{1 + t a n^{2} θ}]$

Differentiating both sides w.r.t. x, we get

$\frac{d y}{d x} = \frac{d}{d x} (2 t a n^{- 1})$

$\frac{d y}{d x} = 2. \frac{1}{1 + x^{2}} = \frac{2}{1 + x^{2}} [∵ \frac{d}{d x} t a n^{- 1} x = \frac{1}{1 + x^{2}}]$

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Find dydxin the following questions: y=sin−1(2x1+x2)

Given, y=sin−1(2x1+x2) putting θ=tan−1x i.e.,x=tan θ Then, y=sin−1(2tant θ1+tan2θ)=sin−1(sin 2θ)=2θ=2tant−1x [∵ sin 2θ=2tantθ1+tan2θ] Differentiating both sides w.r.t. x, we get dydx=ddx(2tan−1) dydx=2.11+x2=21+x2 [∵ ddxtan−1x=11+x2]

Find $\frac{d y}{d x}$ in the following questions:

$y = s i n^{- 1} (\frac{2 x}{1 + x^{2}})$