Given Y=sin(2sin^-1(x)), find dy/dx ?

2 Answers
Jun 26, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(1 - 2 {x}^{2}\right)}{\sqrt{1 - {x}^{2}}}$

Explanation:

Let $t = \arcsin x$, which is defined for $x \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

Then:

$\sin t = x$

and

$\cos t = \sqrt{1 - {x}^{2}}$

because for $x \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the cosine is positive.

So:

$y = \sin \left(2 \arcsin x\right) = \sin 2 t = 2 \sin t \cos t = 2 x \sqrt{1 - {x}^{2}}$

and using the product rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \frac{d}{\mathrm{dx}} \left(\sqrt{1 - {x}^{2}}\right) + 2 \sqrt{1 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \left(- \frac{2 x}{2 \sqrt{1 - {x}^{2}}}\right) + 2 \sqrt{1 - {x}^{2}}$

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(1 - 2 {x}^{2}\right)}{\sqrt{1 - {x}^{2}}}$

Jun 26, 2018

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

${\underbrace{{\sin}^{- 1} y}}_{\alpha} = 2 \underbrace{{\sin}^{-} 1 \left(x\right)} {\setminus}_{\beta} q \quad q \quad \boldsymbol{\alpha = 2 \beta}$

• $\sin \alpha = \sin 2 \beta = 2 \sin \beta \cos \beta$

• $\left\{\begin{matrix}\sin \alpha = y \\ \sin \beta = x \\ \cos \beta = \sqrt{1 - {\sin}^{2} \beta} = \sqrt{1 - {x}^{2}}\end{matrix}\right.$

$\implies y = 2 x \sqrt{1 - {x}^{2}}$

Apply the product rule etc to find:

y'= d/dx(2 x sqrt(1 - x^2)) = 2*(1 - 2 x^2)/sqrt(1 - x^2)