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  • Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 47 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 10  Differentiation  Exercise 10.3 Question 47 Maths Textbook Solution.

Answers (1)

Answer: \frac{2\left ( 6^{x}log6 \right )}{1+6^{2x}}

Hint:

Use substitution method to differentiate this function

Given:

\sin ^{-1}\left ( \frac{2^{x+1}.3^{x}}{1+\left ( 36 \right )^{x}} \right )

Solution:

Let,

\begin{aligned} &y=\sin ^{-1}\left(\frac{2^{x+1} \cdot 3^{x}}{1+(36)^{x}}\right) \\ &\Rightarrow y=\sin ^{-1}\left(\frac{2^{x} 2 \cdot 3^{x}}{1+\left(6^{2}\right)^{x}}\right) \\ &\Rightarrow y=\sin ^{-1}\left(\frac{2 \cdot 6^{x}}{1+(6)^{2 x}}\right) \end{aligned}

Put 6^{x}=\tan \theta

\begin{aligned} &\theta=\tan ^{-1}\left(6^{x}\right) \ldots(i) \\ &\therefore y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \\ &\quad=\sin ^{-1}(\sin 2 \theta) \end{aligned}                                                                                                            \left[\because \sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}\right]

     =2\theta                                                                                                                                            \left[\because \sin ^{-1}(\sin x)=x\right]

     =2\tan ^{-1}\left ( 6^{x} \right )                                                                                                        \left [ using (i) \right ]

Now Differentiating it with respect to x then ,

\begin{aligned} &\frac{d y}{d x}=2 \frac{d}{d x}\left(\tan ^{-1}\left(6^{x}\right)\right) \\ &\qquad=2 \frac{1}{1+\left(6^{x}\right)^{2}} \cdot 6^{\mathrm{x}} \log 6 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \therefore\left[\frac{\mathrm{d}\left(\tan ^{-1} \mathrm{a}^{\mathrm{x}}\right)}{\mathrm{d} \mathrm{x}}=\frac{1}{1+\left(a^{x}\right)^{2}} \frac{\mathrm{d}\left(\mathrm{a}^{\mathrm{x}}\right)}{\mathrm{dx}}\right] \end{aligned}

\begin{aligned} &\quad=\frac{2}{1+6^{2 \mathrm{x}}} \cdot 6^{\mathrm{x}} \log 6 \\ &=\frac{2\left(6^{\mathrm{x}} \log 6\right)}{1+6^{2 \mathrm{x}}} \\ &\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\left(6^{\mathrm{I}} \log 6\right)}{1+6^{2 \mathrm{x}}} \end{aligned}

 

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infoexpert21

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