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

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}