#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 27 Maths Textbook Solution.

Answer:$\frac{dy}{dx}=1$

Hint:

$\frac{d}{dx}\left ( constant \right )=0$

$\frac{d }{dx}\left (x^{n} \right )=nx^{n-1}$

Given:

$\tan ^{-1}\left[\frac{a+\operatorname{btan} x}{b-\operatorname{atan} x}\right]$

Solution:

Let,

\begin{aligned} &y=\tan ^{-1}\left[\frac{a+b \tan x}{b-\operatorname{atan} x}\right] \\ &y=\tan ^{-1}\left[\frac{\frac{a+b \tan x}{b}}{\frac{b-a t a n x}{b}}\right] \\ &y=\tan ^{-1}\left[\frac{\frac{a}{b}+\tan x}{1-\frac{a}{b} \tan x}\right] \end{aligned}

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

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)\right)+\frac{d}{d x}(x) \\ &\frac{d}{d x}(\text { constant })=0 \\ &\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+1 \\ &\frac{d y}{d x}=0+1 \\ &\frac{d y}{d x}=1 \end{aligned}