#### Please solve R.D.Sharma class 12 chapter Differentiation exercise 10.3 question 1 maths textbook solution.

Answer $\frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}$
Hint:

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

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

Solution:

$\begin {array}{ll} Lets\: y=\cos ^{-1}\left\{2 x \sqrt{1-x^{2}}\right\} \\\\ Lets \: \mathrm{x}=\cos \theta \Rightarrow \theta=\cos ^{-1} \mathrm{x} \\\\ Now \ \ y=\cos ^{-1}\left\{2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right\}\end{array}$

$\begin {array}{ll} Using \therefore \cos ^{2} \theta+\sin ^{2} \theta=1,2 \sin \theta \cos \theta=\sin 2 \theta\\\\ \mathrm{y}=\cos ^{-1}\left\{2 \cos \theta \sqrt{\sin ^{2} \theta}\right\}\\\\ \end{array}$

$\begin {array}{ll} y=\cos ^{-1}(2 \cos \theta \sin \theta]\\\\ y=\cos ^{-1}(\sin 2 \theta)\\\\ \therefore \cos \left(\frac{\pi}{2}-\theta\right) \Rightarrow \sin \theta\\\\ y=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\\\ \end{array}$

Now by considering the limits,

$\begin{array}{l} \frac{1}{\sqrt{2}}-2 \theta>-\frac{\pi}{2}\\\\ \Rightarrow \frac{\pi}{2}>\frac{\pi}{2}-2 \theta>\frac{\pi}{2}-\frac{\pi}{2} \end{array}$

Therefore,

$\begin{array}{l} y=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\\\ y=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\\\ y=\left(\frac{\pi}{2}-2 \theta\right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$                                                                                                    $\left\{\cos ^{-1}(\cos \theta)=\theta\right\}$

$y=\frac{\pi}{2}-2 \cos ^{-1} x$

$\text { if } \theta \varepsilon[0, \pi]$
Differentiating with Respect to x, we get

$\begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-2 \cos ^{-1} x\right)\\\\ \therefore \frac{d}{d x}\left(\cos ^{-1} x\right) \Rightarrow \frac{-1}{\sqrt{1-x^{2}}}\\\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}} \text { (constant) }=0\\\\ \frac{d y}{d x}=0-2\left(\frac{-1}{\sqrt{1-x^{2}}}\right)\\\\ \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}} \end{array}$