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  • Explain solution for R.D.Sharma class 12 chapter 10 Differentiation exercise 10.3 question 8 math textbook solution.

Explain solution for R.D.Sharma class 12 chapter 10 Differentiation exercise 10.3 question 8 math textbook solution.

 

Answers (1)

Answer :\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{\sqrt{1-\mathrm{x}^{2}}}

Hint:

\frac{\partial}{\mathrm{d} \mathrm{x}}( Constant )=0 ; \frac{\partial}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-\mathbf{1}}


Given:
\sin ^{-1}\left(1-2 x^{2}\right), 0<x<1

Solution:

\begin {array} {ll}Let \ y=\sin ^{-1}\left\{1-2 x^{2}\right\}\\\\ Let \ x=\sin \theta\\\\ Then,\\ 0=\sin ^{-1} x\\\\ y=\sin ^{-1}\left\{-2 \sin ^{2} \theta\right\}\\\\ Using,\\ 1=2 \sin ^{2} \theta=\cos 2 \theta\\\\ \mathrm{y}=\sin ^{-1}\{\cos 2 \theta\}\\\\ \mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}}{2}-2 \theta\right)\right\} \end{}

As we know,

\begin {array} {ll}\sin \left(\frac{n-\theta}{2}-\theta\right)=\cos \theta \end{}
Considering the limits,

\begin {array} {ll}0<x<1\\\\ 0<\sin \theta<1\\\\ 0<\theta<\frac{n}{2}\ \ \ \ \ \ \ \left\{\sin \frac{1}{2}=1, \sin \theta=0\right\}\\\\ 0 \angle 2 \theta=\pi\\\\ 0>-2 \theta>-\pi\\\\ \frac{1}{2}>\frac{\pi}{2}-2 \theta>-\frac{n}{2}\\\\ \end{}

Now,

\begin {array} {ll}\mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}}{2}-2 \theta\right)\right\} \\\\ y=\frac{n}{2}-2 \theta \\\\ y=\frac{n}{2}=2 \sin ^{-1} x \end{}

Differentiating with respect to x,

\begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left(\frac{n}{2}-2 \sin ^{-1} x\right) \\\\ \text { As we know } \\\\ \frac{d}{d x}(\text { constant })=0 \\\\ \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d y}{d x}=0-2 \frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}} \end{array}

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