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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 18 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 18 Maths Textbook Solution.

Answers (1)

Answer:\frac{d y}{d x}=\frac{2 a^{x} l o g a}{1+a^{2 x}}

Hint:

\frac{d c}{d x}=0;\frac{d}{dx}(x^{n})=nx^{n-1}

Given:

\tan ^{-1}\left\{\frac{2 \mathrm{a}^{x}}{1-\mathrm{a}^{2 x}}\right\}, \mathrm{a}>1,-\infty<\mathrm{x}<0

Solution:

Let

y=\tan ^{-1}\left\{\frac{2 \mathrm{a}^{x}}{1-\mathrm{a}^{2 x}}\right\},

Let,

\begin{aligned} &\begin{aligned} &\mathrm{a}^{\mathrm{x}}=\tan \theta \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\} \\ &\text { Using } \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \\ &\mathrm{y}=\tan ^{-1}(\tan 2 \theta) \end{aligned}\\ &\text { Considering the limits, }\\ &-\infty<\mathrm{x}<0\\ &a^{-\infty}<a^{x}<a^{0} \end{aligned}

0< \tan \theta < 1                                                                                                                \left \{ a^{o} =1\right \}

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

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

y=2\theta                                                                                                        

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

Differentiating with respect to x, we get

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

 

 

 

 

Posted by

infoexpert21

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