Question

2,
45.
If y = sin (2 sin-2x), then (1 - x^2) y, is equal to :

Answer 1

Answer:

Step-by-step explanation:

We have,

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

$\sf{\implies\,y_{1}=cos\left(2\,sin^{-1}(x)\right)\cdot\dfrac{d}{dx}\left(2\,sin^{-1}(x)\right)}$

$\sf{\implies\,y_{1}=cos\left(2\,sin^{-1}(x)\right)\cdot\dfrac{2}{\sqrt{1-{x}^{2}}}}$

$\sf{\implies\,\sqrt{1-{x}^{2}}\cdot\,y_{1}=2\,cos\left(2\,sin^{-1}(x)\right)}$

Differentiating both sides,

$\sf{\implies\,y_{1}\cdot\dfrac{d}{dx}\left(\sqrt{1-{x}^{2}}\right)+\sqrt{1-{x}^{2}}\cdot y_{2}=\dfrac{d}{dx}\left(2\,cos\left(2\,sin^{-1}(x)\right)\right)}$

$\sf{\implies\,y_{1}\cdot\dfrac{-2x}{2\sqrt{1-{x}^{2}}}+\sqrt{1-{x}^{2}}\cdot y_{2}=-2\,sin\left(2\,sin^{-1}(x)\right)\cdot\dfrac{d}{dx}\left(2\,sin^{-1}(x)\right)}$

$\sf{\implies\,-\dfrac{x\cdot y_{1}}{\sqrt{1-{x}^{2}}}+\sqrt{1-{x}^{2}}\cdot y_{2}=-2\,sin\left(2\,sin^{-1}(x)\right)\cdot\dfrac{2}{\sqrt{1-{x}^{2}}}}$

$\sf{\implies\,-x\cdot y_{1}+\left(1-{x}^{2}\right)\cdot y_{2}=-2\,sin\left(2\,sin^{-1}(x)\right)\cdot2}$

$\sf{\implies\,-x\cdot y_{1}+\left(1-{x}^{2}\right)\cdot y_{2}=-4\,sin\left(2\,sin^{-1}(x)\right)}$

$\sf{\implies\,-x\cdot y_{1}+\left(1-{x}^{2}\right)\cdot y_{2}=-4y}$

$\sf{\implies\,\left(1-{x}^{2}\right)\cdot y_{2}=xy_{1}-4y}$