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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 41 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 41 Maths Textbook Solution.

Answers (1)

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

Hint:

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

Given:

\mathrm{y}=\sin \left[2 \tan ^{-1}\left\{\sqrt{\frac{\mathrm{x}}{1+\mathrm{x}}}\right\}\right]

Solution:

Put x=\cos 2\theta

\begin{aligned} &y=\sin \left[2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right] \\ &y=\sin \left[2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}\right] \\ &\text { Since } 1-\cos 2 \theta=2 \sin ^{2} \theta \\ &1+\cos 2 \theta=2 \cos ^{2} \theta \end{aligned}

\begin{aligned} &1+\cos 2 \theta=2 \cos ^{2} \theta \\ &y=\sin \left[2 \tan ^{-1} \sqrt{\tan ^{2} \theta}\right] \\ &\text { Since } \frac{\sin \theta}{\cos \theta}=\tan \theta \\ &y=\sin \left[2 \tan ^{-1}(\tan \theta)\right] \\ &y=\sin [2 \theta] \end{aligned}                                                                            \left \{ \tan ^{-1}\left ( \tan \theta \right )=\theta \: if\: \theta \varepsilon \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] \right \}

y=\sin \left(2 \times \frac{1}{2} \cos ^{-1} x\right)                                                                            \left [ Since,x=\cos 2\theta ,\theta =\frac{1}{2}\cos ^{-1}x \right ]

\begin{aligned} &y=\sin \left(\cos ^{-1} x\right) \\ &y=\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right) \\ &y=\sqrt{1-x^{2}} \end{aligned}                                                                            \left[\sin ^{-1}(\sin \theta) \Rightarrow 0 \text { if } \theta \in\left[\frac{-\pi}{2},\frac{\mathrm{\pi}}{2}\right]\right]

Differentiating it with respect to x

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

As we know,

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


 

Posted by

infoexpert21

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