Question

Please solve this question fasssst....​

Answer 1

Given,

$\longrightarrow\sf{y=\sin^{-1}\left(\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)}$

Its derivative wit respect to x is,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\sin^{-1}\left(\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)\right]\quad\quad\dots(1)}$

Let,

$\longrightarrow\sf{u=\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\quad\quad\dots(i)}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{d}{dx}\left(\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{2}\cdot\dfrac{d}{dx}\left(\sqrt{1+x}+\sqrt{1-x}\right)}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{2}\cdot\dfrac{d}{dx}\left(\sqrt{1+x}\right)+\dfrac{1}{2}\cdot\dfrac{d}{dx}\left(\sqrt{1-x}\right)}$

By chain rule,

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{2}\cdot\dfrac{d(\sqrt{1+x})}{d(1+x)}\cdot\dfrac{d(1+x)}{dx}+\dfrac{1}{2}\cdot\dfrac{d(\sqrt{1-x})}{d(1-x)}\cdot\dfrac{d(1-x)}{dx}}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{2}\cdot\dfrac{1}{2}\,(1+x)^{\frac{1}{2}-1}\cdot(0+1)+\dfrac{1}{2}\cdot\dfrac{1}{2}\,(1-x)^{\frac{1}{2}-1}\cdot(0-1)}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{4}\cdot\dfrac{1}{\sqrt{1+x}}\cdot1+\dfrac{1}{4}\cdot\dfrac{1}{\sqrt{1-x}}\cdot-1}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{4\sqrt{1+x}}-\dfrac{1}{4\sqrt{1-x}}}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{4}\left[\dfrac{1}{\sqrt{1+x}}-\dfrac{1}{\sqrt{1-x}}\right]}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{1}{4}\left[\dfrac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}\right]}$

$\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}\quad\quad\dots(ii)}$

So (1) becomes,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\sin^{-1}u\right)}$

By chain rule,

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{du}\left(\sin^{-1}u\right)\cdot\dfrac{du}{dx}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-u^2}}\cdot\dfrac{du}{dx}}$

From (i) and (ii),

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-\left(\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)^2}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-\left(\dfrac{1+x+1-x+2\sqrt{1-x^2}}{4}\right)}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-\dfrac{1+\sqrt{1-x^2}}{2}}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{\sqrt2}{\sqrt{1-\sqrt{1-x^2}}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{\sqrt2\cdot\sqrt2}{\sqrt2\cdot\sqrt{1-\sqrt{1-x^2}}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{2}{\sqrt{2-2\sqrt{1-x^2}}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-x+1+x-2\sqrt{(1-x)(1+x)}}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{2}{\sqrt{\left(\sqrt{1-x}-\sqrt{1+x}\right)^2}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{4\sqrt{1-x^2}}}$

Since $\sf{\sqrt{1+x}>\sqrt{1-x},}$

$\longrightarrow\sf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1+x}-\sqrt{1-x}}}\cdot\dfrac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^2}}}$

$\longrightarrow\underline{\underline{\sf{\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{1-x^2}}}}}$

Hence 2nd option is the answer.