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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 35 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 35 Maths Textbook Solution.

Answers (1)

Answer: Hence Prove \frac{dy}{dx}=\frac{4}{1+x^{2}}

Hint:

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

Given:

\begin{aligned} &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) \\ &0<x<1 \end{aligned}

Solution:

Prove :\frac{dy}{dx}=\frac{4}{1+x^{2}}

\begin{aligned} &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ &\text { Since } \sec ^{-1} x=\cos ^{-1}\left(\frac{1}{x}\right) \\ &y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}

Let,

    x=\tan \theta

\theta =\tan ^{-1}x

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

Using

\begin{aligned} &\sec ^{2} \theta=1+\tan ^{2} \theta \\ &\frac{1}{\sec \theta}=\cos \theta \\ &\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta \end{aligned}

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right)+\cos ^{-1}(\operatorname{Cos} 2 \theta) \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta)+\cos ^{-1}(\cos 20) \\ &\mathrm{y}=\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \theta)-(\mathrm{i}) \\ &\mathrm{u} \operatorname{sing} 2 \sin \theta \cos \theta=\sin 2 \theta \end{aligned}

Considering Limits,

\begin{aligned} &0<\mathrm{x}<1 \\ &0<\tan \theta<1 \\ &0<\theta<\frac{\mathrm{\pi}}{4} \\ &0<(2 \theta)<\frac{\mathrm{\pi}}{2} \end{aligned}

so from e_{i}\: \: \left ( i \right )                                                                                    

y=2\theta +2\theta

y=4\theta                                                                                                                        

\begin{aligned} &\cos ^{-1}(\cos \theta) \Rightarrow \text { if } \theta \in[0, \pi] \\ &y=4 \tan ^{-1} x \end{aligned}                                                    since,x=\tan \theta ,\theta =\tan ^{-1}x

Differentiating it with respect to x

\begin{aligned} &\frac{d y}{d x}=4 \frac{d}{d x}\left(\tan ^{-1} x\right) \\ &\text { Using } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=4\left(-\frac{1}{1+x^{2}}\right) \\ &\frac{d y}{d x}=\frac{4}{1+x^{2}} \end{aligned}

 

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