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

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}