Question

the derivative of sin inverse (2x √1-x²) w.r.t sin inverse x , 1/√2 <x<1 ,is ​

Answer 1

Answer:

Let u = sin-1 (2x√(1-x2)

Put x = sin θ

Then u = sin-1 (2 sin θ√(1- sin2 θ)

= sin-1 (2 sin θ√cos2 θ

= sin-1 2 sin θ cos θ

= sin-1 sin 2 θ

= 2 θ

= 2 sin-1x

du/dx = 2/√(1-x2)

Let v = sin-1 (3x – 4x3)

Put x = sin θ

v = sin-1 (3sin θ – 4sin3 θ)

= sin-1 sin 3θ

= 3θ

= 3 sin-1 x

dv/dt = 3/√(1-x2)

du/dv = [ 2/√(1-x2) ]/[ 3/√(1-x2) ]

= 2/3

Step-by-step explanation:

I hope it helps you

Answer 2

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

Given functions are

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

and

$\rm :\longmapsto\: {sin}^{ - 1}x$

Let assume that

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

and

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

Now, Consider

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

We use method of Substitution to evaluate this.

So, Substitute

$\rm :\longmapsto\:x = siny$

So, we get

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

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

$\rm :\longmapsto\:u = {sin}^{ - 1}(2siny \: cosy)$

$\rm :\longmapsto\:u = {sin}^{ - 1}(sin2y )$

Now, we know

$\boxed{\tt{ {sin}^{ - 1}(sinx) = x \: \: if \: x \: \in \: \bigg[ - \dfrac{\pi}{2},\dfrac{\pi}{2} \bigg]}}$

Now, given that

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

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{2} } < siny < 1$

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

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

$\rm\implies \: - \dfrac{\pi}{2} > - 2y > - \pi$

$\rm\implies \:\pi - \dfrac{\pi}{2} > \pi - 2y > \pi - \pi$

$\rm\implies \:\dfrac{\pi}{2} > \pi - 2y > 0$

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

Now,

$\rm :\longmapsto\:sin(\pi - 2y) = sin2y$

So,

$\rm :\longmapsto\:u = {sin}^{ - 1}(sin2y )$

can be rewritten as

$\rm :\longmapsto\:u = {sin}^{ - 1}(sin(\pi - 2y ))$

$\rm :\longmapsto\:u = \pi - 2y$

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

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

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

$\rm :\longmapsto\:\dfrac{du}{dx} =\dfrac{d}{dx} \pi - 2 \dfrac{d}{dx}{sin}^{ - 1}x$

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

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

Now, Consider

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

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

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

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

Now, Consider

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

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

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

$\rm \:  =  \: - 2$

Hence,

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

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MORE TO KNOW

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