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  • Please solve RD Sharma class 12 chapter 10 Differentiation exercise 10.3 question 5 maths textbook solution.

Please solve RD Sharma class 12 chapter 10 Differentiation exercise 10.3 question 5 maths textbook solution.

Answers (1)

Answer :   

\frac{\mathrm{d} y}{\mathrm{dx}}=\frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}
Hint:

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

\tan ^{-1}\left\{\frac{x}{\sqrt{a^{2}-x^{2}}}\right\},-a<x<a

Solution:

Let \ \ y=\tan ^{-1}\left\{\frac{x}{\sqrt{a^{2}-x^{2}}}\right\}\\\\

\operatorname{let} \mathrm{x}=\operatorname{asin} \theta\\\\

Now,\\\\

\mathrm{y}=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{\sqrt{a^{2}-a^{2} \\\\\sin ^{2} \theta}}\right\}

\begin{array}{l}Using \ \ \sin ^{2} \theta+\cos ^{2} \theta=1\\\\ y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a \sqrt{1-\sin ^{2} \theta}}\right\}\\\\ y=\tan ^{-1}\frac{a\sin \theta}{a\cos \theta} .\\\\ \mathrm{y}=\tan ^{-1}\{\tan \theta\}\\\\ Using \ \ \frac{\sin \theta}{\cos \theta}=\tan \theta \end{array}
Considering the limits,

\begin{array}{l} -a<x<a\\\\ -a<\operatorname{asin} \theta<\mathbf{a}\\\\ -1<\sin \theta<1\\\\ -\frac{n}{2}<\theta<\frac{n}{2}\ \ \ \ \left\{\sin \frac{n}{2}=1\right\}\\\\ y=\tan ^{-1}(\tan \theta)\\\\ \mathrm{y}=\theta \ \ \ \ \left(\tan ^{-1}(\tan \theta)=(\theta)\right)\) \ \ if \ \ \theta \varepsilon\left[-\frac{n}{2}, \frac{n}{2}\right] \quad \theta=\sin ^{-1}\left(\frac{x}{a}\right) \\\\ Now\\\\ \mathrm{X}=\operatorname{asin} \theta\\\\ \mathrm{y}=\sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right) \end{array}

Differentiating with Respects to x, we get

\begin{array}{l} \frac{d y}{d x}=\frac{d}{\partial x}\left(\sin ^{-1}\left(\frac{x}{a}\right)\right) \\\\ \therefore \frac{\partial}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \times \frac{1}{a} \\\\ \end{array}

\begin{array}{l} \frac{d y}{d x}=\frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a}{\sqrt{a^{2}-x^{2}}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{1}{\sqrt{a^{2}-x^{2}}} \end{array}

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