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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 23 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 23 Maths Textbook Solution.

Answers (1)

Answer:

\frac{dy}{dx}=\frac{2nx^{n-1}}{1+x^{2n}}

Hint:

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

Given:

\begin{aligned} &\cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\} \\ &0<x<\infty \end{aligned}

Solution:

y=cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\}

Let,

x^{n}=\tan \theta

\theta =\tan ^{-1}\left ( x^{n} \right )

y=\cos ^{-1}\left \{ \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } \right \}

y=\cos ^{-1}\left \{ \cos 2\theta \right \}

Using,

\frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta }=\cos 2\theta

Considering the limits,

\begin{aligned} &0<x<\infty \\ &0<x^{n}<\infty \\ &0<\theta<\frac{\pi}{2} \end{aligned}

Now,                                                                                                \left \{ \cos ^{-1}\left ( \cos \theta \right )=\theta ,if\: \theta \varepsilon \left [ 0,\pi \right ] \right \}

y=\cos ^{-1}\left ( \cos 2\theta \right )

y=2\theta                                                                                            Since\hspace{0.2cm}x^{n}=\tan \theta

Differentiating with respect to x we get

\frac{dy}{dx}=\frac{d}{dx}\left ( 2\tan ^{-1}\left ( x^{n} \right ) \right )

Using

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

 

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