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  • Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 13 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 13 Maths Textbook Solution.

Answers (1)

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

Hint:

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

Given:

\begin{aligned} &\tan ^{-1}\left\{\frac{\mathrm{x}}{\mathrm{a}+\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}\right\} \\ &-\mathrm{a}<\mathrm{x}<-\mathrm{a} \end{aligned}

Solution:

Let,

x=a\sin \theta

\theta =\sin ^{-1}\frac{x}{a}

Now,

\begin{aligned} &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+a \sqrt{1-\sin ^{2} \theta}}\right\} \end{aligned}

Using

\begin{aligned} &\cos ^{2} \theta+\sin ^{2} \theta=1 \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+a \sqrt{\cos \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a+\operatorname{acos} \theta}\right\} \\ &y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a(1+\operatorname{los} \theta)}\right\} \\ &U \operatorname{sing} 2 \cos ^{2} \theta=1+\cos \theta \\ &y=\tan ^{-1}\left\{\frac{\sin \theta}{1+\cos \theta}\right\} \text { and } \\ &2 \sin \theta \cos \theta=\sin 2 \theta \\ &y=\tan ^{-1}\left\{\frac{\sin \theta}{1+\cos \theta}\right\} \end{aligned}

\begin{aligned} &y=\tan ^{-1}\left\{\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}\right\} \\ &y=\tan ^{-1}\left\{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right\} \\ &y=\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \end{aligned}                                                    \therefore \frac{\sin \theta }{\cos \theta }=\tan \theta

Considering the limit

\begin{aligned} &-a<x<a \\ &-1<\sin \theta<1 \\ &\frac{-\pi}{2}<\theta<\frac{\pi}{2} \\ &-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4} \end{aligned}                                                                            \left \{ \sin \frac{\pi}{2}=1 \right \}

Now,

\begin{aligned} &y==\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \\ &y=\frac{\theta}{2} \\ &y=\frac{1}{2} \sin ^{-1} \frac{x}{a} \end{aligned}

Differentiating with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{2} \sin ^{-1} \frac{x}{a}\right) \\ &\therefore \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{2} \frac{1}{\sqrt{1-(x)^{2}}} \times \frac{1}{a} \\ &\Rightarrow \frac{1}{2} \times \frac{1}{\sqrt{1-\frac{x^{2}}{a^{2}}}} \times \frac{1}{a} \\ &\Rightarrow \frac{d y}{d x}=\frac{1}{2} \times \frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}} \times \frac{1}{a} \\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \end{aligned}

 

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