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  • Please solve R.D.Sharma class 12 chapter 10 Differentiation exercise 10.3 question 9 maths textbook solution.

Please solve R.D.Sharma class 12 chapter 10 Differentiation exercise 10.3 question 9 maths textbook solution.

Answers (1)

Answer :  

\frac{d y}{d x}=\frac{-a}{a^{2}+x^{2}}

Hint:

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

Given:

\cos ^{-1}\left\{\frac{x}{\sqrt{x^{2}+a^{2}}}\right\}

Solution:

Let,\\\\

\mathrm{y}=\cos ^{-1}\left\{\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)\\\\

Let\\\\

\mathrm{x}=\operatorname{acot} \theta\\\\

\theta=\cot ^{-1}\left(\frac{x}{a}\right)\\\\

\begin{array}{l} \text { Now }\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{\sqrt{a^{2} \cot ^{2} \theta+a^{2}}}\right\}\\\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{\sqrt{a^{2}\left(\cot ^{2} \theta+1\right)}}\right\}\\\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{a \sqrt{\cot ^{2} \theta+1}}\right\}\\\\ \text { Using } 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\\\\ y=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{a \sqrt{\operatorname{cosec}^{2} \theta}}\right\}\\\\ \mathrm{y}=\cos ^{-1}\left\{\frac{\operatorname{acot} \theta}{\mathrm{acosec} \theta}\right\}\\\\ \mathrm{y}=\cos ^{-1}\left\{\frac{\cot \theta}{\operatorname{cosec} \theta}\right\} \end{array}

As we Know,

\begin{array}{l} \cot \theta=\frac{\cos \theta}{\sin \theta} \text { , }\\\\ \operatorname{cosec} \theta=\frac{1}{\sin \theta}\\\\ \mathrm{y}=\cos ^{-1}\left\{\frac{\cos \theta}{\sin \theta} \times \sin \theta\right\}\\\\ y=\cos ^{-1}(\cos \theta)\\\\ \mathrm{y}=\theta\\\\ \mathrm{y}=\cot ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\\\\ \text { Differentiating with respect to } x \text { We get }\\\\ \frac{d y}{d x}=\frac{d}{d x}\left(\cot ^{-1}\left(\frac{x}{a}\right)\right)\\\\ \therefore \frac{\partial}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{1+x^{2}}\\\\ \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{-1}{1+\left(\frac{\mathrm{x}}{\mathrm{a}}\right)^{2}} \times \frac{1}{\mathrm{a}} \end{array}

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

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