Question

differentiate the following ​.
jo koi solve karega use

Answer 1

Solution :-

We have

$\sf{ y = sin^{-} ( 2x\sqrt{1 - x^2})}$

Now in order to solve it let us put

$\sf{ x = sin\theta }$

$\implies \sf{ \theta = sin^{-}x}$

By using x = sin∅

$\implies \sf{ y = sin^{-} ( 2sin\theta\sqrt{1 - \sin^2\theta})}$

$\implies \sf{ y = sin^{-} ( 2sin\theta\sqrt{cos^2\theta})}$

$\implies \sf{ y = sin^{-} ( 2sin\theta cos\theta)}$

As 2sin∅cos∅ = sin2∅

$\implies \sf{ y = sin^{-} ( sin2\theta)}$

$\implies \sf{ y = 2\theta}$

or

$\implies \sf{ y = 2sin^{-} x }$

Now differentiate y w.r.t. x

$= \sf{ \dfrac{dy}{dx} }$

$= \sf{ \dfrac{d(2 sin^{-}x)}{dx} }$

$= \sf{ 2.\dfrac{d(sin^{-}x)}{dx} }$

Now as

$\sf{\bold{ \dfrac{d(sin^{-}x)}{dx}= \dfrac{1}{\sqrt{1 - x^2 }} }}$

$= \sf{ 2 \dfrac{1 }{\sqrt{1 - x^2}} }$

$= \sf{ \dfrac{2 }{\sqrt{1 - x^2}} }$

Answer 2

Answer:

$y = \sin {}^{ - 1} (2x \sqrt{1 - x {}^{2} } )$

let x = sin ß

thus

$\sin {}^{ - 1} (x) = \beta$

therefore ;

$y = \sin {}^{ - 1} (2 \sin( \beta ) \sqrt{1 - \sin {}^{2} ( \beta ) } )$

$y = \sin {}^{ - 1} (2 \sin( \beta ) \cos( \beta) )$

$y = \sin {}^{ - } ( \sin(2 \beta ) )$

$y = 2 \beta$

$y = 2 \sin {}^{ - 1} (x)$

now differnatiate with respect to x

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

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