Question

The derivative of $\sf {sin-1 \left(2x \sqrt{(1 - x2)} )\right$ w.r.t $\sf{sin-1 x, { \dfrac{1}{\sqrt{2}} < x < 1}$ ,is ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given to differentiate

$\rm \: {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) \: w.r.t \: \: {sin}^{ - 1}x$

Let assume that

$\rm \: u \: = \: {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) \:$

and

$\rm \: v = {sin}^{ - 1}x$

Now, Consider

$\rm \: u \: = \: {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) \:$

Let we substitute,

$\rm \: x = sin \theta \:$

So, above expression can be rewritten as

$\rm \: u = {sin}^{ - 1}(2sin\theta \sqrt{1 - {sin}^{2}\theta } )$

$\rm \: u = {sin}^{ - 1}(2sin\theta \sqrt{{cos}^{2}\theta } )$

$\rm \: u = {sin}^{ - 1}(2sin\theta \: cos\theta )$

$\rm \: u = {sin}^{ - 1}(sin2\theta )$

Given that,

$\rm \: \dfrac{1}{ \sqrt{2} } < x < 1$

$\rm \: \dfrac{1}{ \sqrt{2} } < sin\theta < 1$

$\rm\implies \:\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2}$

$\rm\implies \:\dfrac{\pi}{2} <2 \theta < \pi$

$\rm\implies \: - \pi < - 2 \theta < - \dfrac{\pi}{2}$

$\rm\implies \: \pi- \pi <\pi - 2 \theta < \pi - \dfrac{\pi}{2}$

$\rm\implies \: 0 <\pi - 2 \theta < \dfrac{\pi}{2}$

So, above expression

$\rm \: u = {sin}^{ - 1}(sin2\theta )$

can be rewritten as

$\rm \: u = {sin}^{ - 1}[sin(\pi - 2\theta )]$

We know,

$\boxed{\sf{ \:{sin}^{ - 1}(sinx) =   x \:  \: if -\dfrac{\pi  }{2} \leqslant x \leqslant \dfrac{\pi  }{2} \: }}$

So, using this, we get

$\rm \: u = \pi - 2\theta$

$\rm \: u = \pi - 2{sin}^{ - 1}x$

$\rm \: [ \: \because \: x = sin\theta \: \rm\implies \:\theta = {sin}^{ - 1}x \: ]$

On differentiating w. r. t. x, we get

$\rm \: \dfrac{d}{dx}u \: = \: \dfrac{d}{dx}(\pi - 2{sin}^{ - 1}x)$

$\rm \: \dfrac{du}{dx} \: = \: 0 - \dfrac{2}{ \sqrt{1 - {x}^{2} } }$

$\rm\implies \:\boxed{\sf{ \: \: \rm \: \dfrac{du}{dx} \: = \: - \dfrac{2}{ \sqrt{1 - {x}^{2} } } \: }}$

Now, Consider

$\rm \:v \: = \: {sin}^{ - 1}x$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}v \: = \: \dfrac{d}{dx}{sin}^{ - 1}x$

$\rm \: \dfrac{dv}{dx} \: = \: \dfrac{1}{ \sqrt{1 - {x}^{2} } }$

Now, Consider

$\rm \: \dfrac{du}{dv}$

$\rm \: = \: \dfrac{du}{dx} \div \dfrac{dv}{dx}$

$\rm \: = \: - \: \dfrac{2}{ \sqrt{1 - {x}^{2} } } \div \dfrac{1}{ \sqrt{1 - {x}^{2} } }$

$\rm \: = \: - 2$

Hence,

$\rm\implies \:\rm \: \dfrac{du}{dv} \: = \: - \: 2$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y =  {sin}^{ - 1}(sinx) & \sf  x \:  \: if -\dfrac{\pi  }{2} \leqslant x \leqslant \dfrac{\pi  }{2}\\ \\ \sf y =  {cos}^{ - 1}(cosx) & \sf x \:  \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y =  {tan}^{ - 1}(tanx) & \sf x \:  \: if \:  - \dfrac{\pi  }{2} < x < \dfrac{\pi  }{2}\\ \\ \sf y =  {cosec}^{ - 1}(cosecx) & \sf x \:  \: if \: x \:  \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi  }{2}\bigg] -  \{0 \}\\ \\ \sf y =  {sec}^{ - 1}(secx) & \sf x \:  \: if \: x \:  \in \: [0, \: \pi] \:   -  \: \bigg\{\dfrac{\pi  }{2}\bigg\}\\ \\ \sf y =  {cot}^{ - 1}(cotx) & \sf x \:  \: if \:  \:  \in \: \bigg( -  \dfrac{\pi  }{2} , \dfrac{\pi  }{2}\bigg) -  \{0 \} \end{array}} \\ \end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

I hope it will help you.

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