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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 28 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 28 Maths Textbook Solution.

Answers (1)

Answer:\frac{dy}{dx}=\frac{1}{1+x^{2}}

Hint:

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

Given:

\tan ^{-1}\left ( \frac{a+bx}{b-ax} \right )

Solution:

Let,

\begin{aligned} &y=\tan ^{-1}\left(\frac{a+b x}{b-a x}\right) \\ &y=\tan ^{-1}\left[\frac{\frac{a+b x}{b}}{\frac{b-a x}{b}}\right] \end{aligned}

 

Since,\tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\left ( \frac{x+y}{1-xy} \right )

Differentiating it with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1} x\right. \\ &\text { Since, } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+\left(\frac{a}{b}\right)^{2}} \times 0+\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=0+\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+x^{2}} \end{aligned}

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