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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 16 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 16 Maths Textbook Solution.

Answers (1)

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

Hint:

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

Given:

\begin{aligned} &\tan ^{-1}\left\{\frac{4 x}{1-4 x^{2}}\right\} \\ &\frac{-1}{2}<x<\frac{1}{2} \end{aligned}

Solution:

Let,

y=\tan ^{-1}\left\{\frac{4 x}{1-4 x^{2}}\right\}

Let,

2x=\tan \theta

\theta =\tan ^{-1}2x

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

Using \tan 2\theta =\frac{2\tan \theta }{1-\tan ^{2}\theta }

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

Considering the limits,

\begin{aligned} &\frac{-1}{2}<x<\frac{1}{2} \\ &-1<2 x<1 \\ &-1<\tan \theta<1 \\ &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}                       \left \{ \tan 45^{\circ}=1 \right \}

Now,

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

y=2\theta                                                            

y=2\tan ^{-1}\left ( 2x \right )                                                        \left [ since,2x=\tan \theta \right ]

Differentiating with respect to x , We get

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

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