#### Need solution for R.D.Sharma maths class 12 chapter 10 Differentiation exercise 10.3 question 3 math Textbook solution.

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

Hint:

Use substitution method to differentiate this function

Given:

$\sin ^{-1}\left\{\sqrt{ \left.\frac{1-x}{2}\right\}}\right\}, 0

Solution:

$\begin{array}{l} \operatorname{Let} y=\sin ^{-1}\left\{\sqrt{\frac{1-x}{2}}\right\}\\\\ \end{array}$

$\begin{array}{l} \text { let } x =\cos 2 \theta \\\\ \left[\theta=\frac{1}{2} \cos ^{-1} x\right] \ \ - (1) \\ \end{array}$

$\begin{array}{l} \therefore y=\sin ^{-1}\left\{\sqrt{\frac{1-\cos 2 \theta}{2}}\right\} \\\\ =\sin ^{-1}\left\{\sqrt{\frac{2 \sin ^{2} \theta}{2}}\right\} \\\\ y =\sin ^{-1}\left\{\sqrt{\sin ^{2} \theta}\right\} \ \ \ \ \left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right] \\\\ =\sin ^{-1}(\sin \theta) \\\\ y =\theta=\frac{1}{2} \cos ^{-1} x[\text { using (i)] } \end{array}$

Now differentiating y w.r.t x then

$\begin{array}{l} \frac{d y}{d x}=\frac{1}{2} \frac{d\left(\cos ^{-1} x\right)}{d x} \\\\ \quad=\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^{2}}} \\\\ \quad=\frac{-1}{2 \sqrt{1-x^{2}}} \quad\left[\because \frac{d}{d x} \cos ^{-1} x=\frac{-1}{\sqrt{1-x^{2}}}\right] \\\\ \therefore \frac{d y}{d x}=\frac{-1}{2 \sqrt{1-x^{2}}} \end{array}$