Find d y-d x- if y-sin -1 x-sin -1-1-x2-1 - x- - 1-

Given, y=sin^ 1x+sin^ 1√(1 x^2) Putting sin^ 1x=θ ⇒ x=sinθ ⇒ y=θ+sin^ 1√(1 sin^2θ)⇒ y=θ+sin^ 1(cos θ) [∵ 1 sin^2θ=cos^2θ] ⇒ y=θ+sin^ 1sin(π/2 θ )⇒ y=θ+π/2 ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

Find d y/d x,...

Question

$F i n d \frac{d y}{d x}, i f y = s i n^{- 1} x + s i n^{- 1} \sqrt{1 - x^{2}}, - 1 \leq x, \leq 1.$

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Solution

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

$P u t t i n g s i n^{- 1} x = θ \Rightarrow x = s i n θ \Rightarrow y = θ + s i n^{- 1} \sqrt{1 - s i n^{2} θ} \Rightarrow y = θ + s i n^{- 1} (c o s θ) [∵ 1 - s i n^{2} θ = c o s^{2} θ] \Rightarrow y = θ + s i n^{- 1} s i n (\frac{π}{2} - θ) \Rightarrow y = θ + \frac{π}{2} - θ = \frac{π}{2} (∴ s i n (\frac{π}{2} - θ) = c o s θ) \Rightarrow y = \frac{π}{2}$

On differentiation w.r.t. x, we get $\frac{d y}{d x} = 0$

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Find dydx, if y=sin−1x+sin−1√1−x2,−1≤x,≤1.

Given, y=sin−1x+sin−1√1−x2 Putting sin−1x=θ ⇒ x=sinθ⇒ y=θ+sin−1√1−sin2θ⇒ y=θ+sin−1(cos θ) [∵ 1−sin2θ=cos2θ]⇒ y=θ+sin−1sin(π2−θ)⇒ y=θ+π2−θ=π2(∴ sin(π2−θ)=cos θ)⇒ y=π2 On differentiation w.r.t. x, we get dydx=0

$F i n d \frac{d y}{d x}, i f y = s i n^{- 1} x + s i n^{- 1} \sqrt{1 - x^{2}}, - 1 \leq x, \leq 1.$