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

Answers (1)

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<x<\frac{1}{2}

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<x=1 \\ &0<2 x<1 \\ &0<\cos \theta=1 \\ &0<\theta<\frac{\pi}{2} \\ &\text { And } 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}



 

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