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

Answers (1)

Answer: $\frac{d y}{d x}=1$

Hint:

$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}} \text { (constants) }=0$

$\frac{d}{d\mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$

Given:

$\sin ^{-1}\left\{\frac{\sin x+\cos x}{\sqrt{2}}\right\}$

$\frac{-3 \mathrm{\pi}}{4}<\mathrm{x}<\frac{\pi}{4}$

Soluton:

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

Now,

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

$\mathrm{y}=\sin ^{-1}\left\{\sin \mathrm{x} \cos \left(\frac{\mathrm{\pi}}{4}\right)+\cos \mathrm{xsin}\left(\frac{\mathrm{\pi}}{4}\right)\right\}$

$\sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \&$

$\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$

Using,

$\sin (A+B)=\sin A \cos B+\cos A \sin B$

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

Considering the limits,

$-\frac{3 \pi}{4}<x<\frac{\pi}{4}$

$-\frac{3 \pi}{4}+\frac{\pi}{4}<x+\frac{\pi}{4}<\frac{\pi}{4}+\frac{1}{4}$

$-\frac{\mathrm{\pi}}{2}<\mathrm{x}+\frac{\mathrm{\pi}}{4}<\frac{\pi}{2}$

Now,

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

$y=x+\frac{\pi}{4}$ $\left\{\sin ^{-1}(\sin \theta)=\theta, \text { if } \theta \varepsilon\left[\frac{\pi}{2}, \frac{\pi}{2}\right]\right\}$

Differentiating with respect to x , We get

$\frac{d y}{d x}=1+0$

$\frac{\mathrm{d}}{\mathrm{dx}} \text { (constant) }=0$

$\frac{d y}{d x}=1$

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