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

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

Answers (1)

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}}<x<\frac{1}{3 \sqrt{2}} \end{aligned}

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}

Posted by

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