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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 44 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 44 Maths Textbook Solution.

Answers (1)

Answer:\frac{dy}{dx}=\frac{-6}{\sqrt{1-4x^{2}}}

Hint:

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

Given:

 

Solution:

\begin{aligned} &y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}}, \\ &-\frac{1}{2}<x<0 \end{aligned}

Put2x=\cos \theta

So

y=\cos ^{-1}(\operatorname{coc} \theta)+2 \cos ^{-1} \sqrt{1-\cos ^{2} \theta}

Using,\sin ^{2}\theta +\cos ^{2}\theta =1

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

Considering limit,

\begin{aligned} &-\frac{1}{2}<\mathrm{x}<0 \\ &-1<2 \mathrm{x}<0 \\ &-1-\cos \theta<0 \\ &\frac{\mathrm{\pi}}{2}<\theta<\mathrm{\pi} \end{aligned}

And,

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

So, from equation (i)

\mathrm{y}=\theta+2\left[-\left(\frac{\pi}{2}-\theta\right)\right]

                               \left\{\operatorname{Since}, \cos ^{-1}(\cos \theta)=0 \text { if } \theta \varepsilon[0, \pi], \cos ^{-1} \cos (\theta)=-\theta, \text { if } \theta \varepsilon[-\pi \cdot 0]\right\}

\begin{aligned} &\mathrm{y}=\theta-2 \times \frac{\pi}{2}+2 \theta \\ &\mathrm{y}=-\pi+3 \theta \\ &\mathrm{y}=-\pi+3 \cos ^{-1}(2 \mathrm{x}) \end{aligned}

Differentiating its with respect to x using chain rule

\frac{d y}{d x}=0+3\left[\frac{-1}{\sqrt{1-(2 x)^{2}}}\right] \frac{d}{d x}(2 x)

As we Know,

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-3}{\sqrt{1-4 \mathrm{x}^{2}}}-(2) \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{6}{\sqrt{1-4 \mathrm{x}^{2}}} \end{aligned}

Posted by

infoexpert21

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