#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 42 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{2}{\sqrt{1-4x^{2}}}$

Hint:

\begin{aligned} &\frac{d}{d \mathrm{x}} \text { (constant) }=0 \\ &\frac{d}{d\mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{-1} \end{aligned}

Given:

$y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1+4 x^{2}}, 0

Solution:

Put $2x=\cos \theta$

So,

\begin{aligned} &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1} \sqrt{1-\cos ^{2} \theta} \\ &\text { Since } \cos ^{2} \theta+\sin ^{2} \theta=1 \\ &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1} \sqrt{\sin ^{2} \theta} \\ &y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1}(\sin \theta) \\ &y=\cos (\cos \theta)+2 \cos ^{-1}\left(\cos \left(\frac{\pi}{2}-\theta\right)\right)-(i) \end{aligned}

\begin{aligned} &0-\theta>-\frac{\pi}{2} \\ &\frac{n}{2}>\left(\frac{n}{2}-\theta\right)>0 \end{aligned}

So From eq (i)

$y=\theta +2\left ( \frac{\pi}{2}-\theta \right )$                                                                                $\left \{ Since \cos ^{-1}\left ( \cos \theta \right )=0\: if\: \theta \varepsilon \left [ 0,n \right ] \right \}$

\begin{aligned} &y \Rightarrow \theta+\pi-2 \theta \\ &y \Rightarrow \pi-C \\ &y=\pi-\cos ^{-1}(2 x) \\ &\text { Since, } 2 x=\cos \theta \end{aligned}

Differentiating it with respect to $x$

$\frac{d y}{d x}=0-\left[\frac{-1}{\sqrt{1-(2 x)^{2}}}\right] \frac{d}{d x}(2 x)$

Since,$\frac{d}{dx}\left ( constant \right )=0;$

\begin{aligned} &\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{\sqrt{1-4 x^{2}}}(2) \\ &\frac{d y}{d x}=\frac{2}{\sqrt{1-4 x^{2}}} \end{aligned}