#### Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3  Question 14 Maths Textbook Solution.

Answer:$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}$

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{d}{d x}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}

Given:

\begin{aligned} &\sin ^{-1}\left\{\frac{\mathrm{x}+\sqrt{1-\mathrm{x}^{2}}}{\sqrt{2}}\right\} \\ &-1<\mathrm{x}<1 \end{aligned}

Solution:

Let

$y=\sin ^{-1}\left\{\frac{\mathrm{x}+\sqrt{1-\mathrm{x}^{2}}}{\sqrt{2}}\right\}$

Let

$x=\sin \theta$

$\theta =\sin ^{-1}x$

Now,

$\mathrm{y}=\sin ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\}$

Using

\begin{aligned} &\sin ^{2} \theta+\cos ^{2} \theta=1 \\ &y=\sin ^{-1}\left\{\frac{\sin \theta+\sqrt{\cos ^{2} \theta}}{\sqrt{2}}\right] \\ &y=\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\} \end{aligned}

Now,

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left\{\sin \theta \frac{1}{\sqrt{2}}+\cos \theta \frac{1}{\sqrt{2}}\right\} \\ &\mathrm{y}=\sin ^{-1}\left\{\sin \theta \cos \left(\frac{\mathrm{\pi}}{4}\right)+\cos \theta \sin \left(\frac{\mathrm{\pi}}{4}\right)\right\} \\ &\sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \& \cos \left(\frac{\mathrm{\pi}}{4}\right) \end{aligned}

$y=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$

Considering the limits,

\begin{aligned} &-1                                                $\left \{ \sin \frac{\pi}{2}=1 \right \}$

\begin{aligned} &\frac{-\pi}{2}+\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{\pi}{2}+\frac{\pi}{4} \\ &-\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{3 \pi}{4} \end{aligned}

Now,

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left\{\sin \left(\theta+\frac{\mathrm{\pi}}{4}\right)\right\} \\ &\mathrm{y}=\theta+\frac{\mathrm{\pi}}{4} \\ &\mathrm{y}=\sin ^{-1} \mathrm{x}+\frac{\mathrm{\pi}}{4} \end{aligned}                        $sin^{-}\left ( \sin \theta \right )=\theta if\theta \varepsilon \left [ \frac{-\pi}{2} ,\frac{\pi}{2}\right ]$Differentiating with respect to x , We get

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \\ &\frac{\mathrm{d}}{d \mathrm{x}} \text { (constant) }=0 \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \end{aligned}