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

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

Answers (1)

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<x<1 \\ &-1<\sin \theta<1 \\ &-\frac{\pi}{2}<\theta<\frac{\pi}{2} \end{aligned}                                                \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}

 

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