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  • Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 20 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3  Question 20 Maths Textbook Solution.

Answers (1)

Answer: \frac{dy}{dx}=\frac{a}{2\left ( 1+a^{2}x^{2} \right )}

Hint:

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

Given:

\begin{aligned} &\tan ^{-1}\left\{\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right\}, \\ &x \neq 0 \end{aligned}

Solution:

Let,

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

Let\: ax=\tan \theta

Now

y=\tan ^{-1}\left \{ \frac{\sqrt{1+\tan ^{2}\theta -1}}{\tan \theta } \right \}

Using

\begin{aligned} &\sec ^{2} \theta=1+\tan ^{2} \theta \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sec \theta-1}{\tan \theta}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right\} \end{aligned}

\begin{aligned} &\text { Using } \tan \theta=\frac{\sin \theta}{\cos \theta}, \sec \theta=\frac{1}{\cos \theta} \\ &y=\tan ^{-1}\left\{\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin d}{\cos \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right\} \\ &y=\tan ^{-1}\left\{\frac{1-\cos \theta}{\sin \theta}\right\} \end{aligned}

\begin{aligned} &\text { Using } 2 \sin ^{2} \theta=1-\cos 2 \theta \\ &\text { and } 2 \sin \theta \cos \theta=\sin 2 \theta \\ &y=\tan ^{-1}\left\{\frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2}\right\} \\ &y=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \end{aligned}

y=\frac{\theta }{2}                                                                                            

y=\frac{1}{2}tan^{-1}ax                                                                        

Differentiating with respect to x , We get

\begin{aligned} &\frac{dy}{d x}=\frac{d}{d x}\left(\frac{1}{2} \operatorname{tan}^{-1} x\right)\\ &\text { Using }\\ &\frac{d}{dx}\left(\operatorname{tan}^{-1} x\right) \Rightarrow \frac{1}{1+x^{2}}\\ &\frac{d y}{d x}=\frac{1}{2} \times \frac{1}{1+(a x)^{2}} \times \frac{1}{a}\\ &\frac{d y}{d x}=\frac{a}{2\left(1+a^{2} x^{2}\right)} \end{aligned}

 

 

Posted by

infoexpert21

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