#### Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3  Question 48 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{6}{\sqrt{1-9x^{2}}}$

Hint:

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

Given:

\begin{aligned} &y=\sin ^{-1}\left(6 x \sqrt{1-9 x^{2}}\right) \\ &\frac{1}{3 \sqrt{2}}

Solution:

Let

\begin{aligned} &\mathrm{x}=\frac{1}{3} \sin \theta \\ &\mathrm{y}=\sin ^{-1}\left(6 \times \frac{1}{3} \sin \theta \sqrt{\left.1-9\left(\frac{1}{3}\right)^{2} \sin ^{2} \theta\right)}\right. \\ &\mathrm{y}=\sin ^{-1}\left(2 \sin \theta \sqrt{1-9 \times \frac{1}{9} \sin ^{2} \theta}\right) \end{aligned}

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\ &\mathrm{U} \sin g \sin ^{2} \theta+\cos ^{2} \theta=1 \\ &\mathrm{y}=\sin ^{-1}\left(2 \sin \theta \sqrt{\cos ^{2} \theta}\right) \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta) \\ &\mathrm{y}=\sin ^{-1}(\sin 2 \theta)-(\mathrm{i}) \\ &\mathrm{U} \sin g \sin 2 \theta=2 \sin \theta \cos \theta \end{aligned}

Considering limits here

\begin{aligned} &\frac{1}{3 \sqrt{2}}<\mathrm{x}<\frac{1}{3 \sqrt{2}} \\ &\frac{1}{3} \times \frac{1}{3 \sqrt{2}}<\frac{1}{3} \sin \theta<\frac{1}{3 \sqrt{2}} \times \frac{1}{3} \\ &\frac{1}{9 \sqrt{2}}<\frac{1}{3} \sin \theta<\frac{1}{9 \sqrt{2}} \end{aligned}

From Equation (i)

$y=2\theta$                                                                                                                                $\left\{\begin{array}{l} \sin ^{-1}(\sin \theta)=\theta \\ \text { if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \end{array}\right\}$

$y=2\sin ^{-1}\left ( 3x \right )$

Differentiating it with respect to $x$

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \sin ^{-1} 3 x\right) \\ &\text { As we thow, } \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=2 \times \frac{1}{\sqrt{1-(3 x)^{2}}} \frac{d}{d x}(3 x) \\ &\frac{d y}{d x}=\frac{2}{\sqrt{1-9 x^{2}}} \times 3 \\ &\frac{d y}{d x}=\frac{6}{\sqrt{1-9 x^{2}}} \end{aligned}