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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Diffrentiation Exercise 10.3 Question 17 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Diffrentiation Exercise 10.3 Question 17 Maths Textbook Solution.

Answers (1)

Answer: \frac{dy}{dx}=\frac{2^{x+1}log2}{1+4x}

Hint:

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

Given:

\begin{aligned} &\tan ^{-1}\left[\frac{2^{x+1}}{1-4^{x}}\right] \\ &-\infty<x<0 \end{aligned}

Solution:

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

Let,

2^{x}=\tan \theta

y=\tan ^{-1}\left \{ \frac{2x2^{x}}{1-\left ( 2x \right )^{2}} \right \}

As we Know

\begin{aligned} &a^{m+n} \rightarrow a^{m} \cdot a^{n} \\ &y=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\} \end{aligned}

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} &-\infty<x<0 \\ &2^{-\infty}<2^{x}<2^{0} \\ &0<\tan \theta<1 \\ &0<\frac{\pi}{4} \end{aligned}                                                    \left\{2^{0}=1,2^{-\infty}=0\right\}

0< 2\theta < \frac{\pi}{2}                                                                \left \{ \tan \frac{\pi}{4} =1\right \}

0< 2\theta < \frac{\pi}{2}                                               

Now

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

y==2\theta

y=2\tan ^{-1}\left ( 2^{x} \right )                                                                                                         \left\{\begin{array}{r} \sin c e 2^{x}=\tan \theta \\ \theta=\tan ^{-1}\left(2^{x}\right) \end{array}\right\}

Differentiating with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1}\left(2^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{2^{x} \log 2}{1+\left(2^{x}\right)^{2}} \\ &\frac{d y}{d x}=\frac{2^{x+1} \log 2}{1+4^{x}} \end{aligned}                                                                            \begin{gathered} \therefore \frac{\mathrm{d}\left(\tan ^{-1} \mathbf{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(2^{\mathrm{x}}\right)=\log 2 \\ \left\{\mathrm{a}^{\mathrm{m}} \cdot \mathbf{a}^{\mathrm{n}} \Rightarrow \mathbf{a}^{\mathrm{m}+\mathrm{h}}\right\} \end{gathered}

 

   

                                                                                         

                         

Posted by

infoexpert21

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