#### Provide solution for R.D.Sharma math class 12 chapter Differentiation exercise 10.3 question 2 Maths Textbook solution.

Answer :  $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right)$

Hint:

$\begin {array} {ll}\cos ^{-1}(\cos \theta)=\theta \quad if \ \ \theta \varepsilon[0, \mathrm{\pi}] \\\\ \cos ^{-1}(\cos \theta)=-\theta \quad if \ \ \ \theta \varepsilon[-\mathrm{\pi}, 0] \end{array}$
Given:

$\begin {array} {ll}\cos ^{-1}\left\{\sqrt{\frac{1+x}{2}}\right\},-1

Solution:

$\begin {array} {ll}Let, y=\cos ^{-1}\left\{\sqrt{\frac{1+x}{2}}\right\} \\\\ Let \: \,x=\cos 2 \theta, \theta=\frac{1}{2} \cos ^{-1} x \\\\ Now, y=\cos ^{-1}\left\{\sqrt{\frac{1+\cos 2 \theta}{2}}\right\}\end{array}$

$\begin {array} {ll}By \ using\ \cos 2 \theta=2 \cos ^{2} \theta-1\\\\ y=\cos ^{-1}\left\{\sqrt{\frac{2 \cos ^{2} \theta}{2}}\right\}\\\\ y=\cos ^{-1}(\cos \theta)\\\\ \therefore \cos ^{-1}(\cos \theta)=\theta\\\\\end{array}$

$\begin {array} {ll} Now\\\\ y=\cos ^{-1}(\cos \theta)\\\\ y=\theta\\\\ y=\frac{1}{2} \cos ^{-1} x \ \ \ \ \ \ \ \therefore \cos ^{-1}(\cos \theta)=\theta \quad if \theta \varepsilon[0, \pi] \end{array}$

Differentiating with respect to x, we get

$\begin {array} {ll} \frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow \frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right) \end{array}$

As we know,

$\begin {array} {ll} \frac{\mathrm{d}}{\mathrm{dx}} \ (constants) =0 \\\\ \frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}} \end{array}$

Hence, final answer is $\begin {array} {ll} \frac{1}{2}\left(\frac{-1}{\sqrt{1-x^{2}}}\right) \end{array}$