#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 31 Maths Textbook Solution.

Answer:$\frac{dy}{dx}=\frac{3}{1+9x^{2}}+\frac{2}{1+4x^{2}}$

Hint:

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

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

Given:

$\tan ^{-1}\left ( \frac{5x}{1-6x^{2}} \right )$

Solution:

Let,

\begin{aligned} &y=\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right) \\ &y=\tan ^{-1}\left(\frac{3 x+2 x}{1-(3 x)(2 x)}\right) \end{aligned}

\begin{aligned} &\text { By using } 5 x=3 x+2 x \\ &y=\tan ^{-1}(3 x)+\tan ^{-1}(2 x) \end{aligned}

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

Differentiating it with respect to x using chain rule,

$\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}(3 x)\right)+\frac{d}{d x}\left(\tan ^{-1}(2 x)\right)$

Using

\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{1}{1+(3 x)^{2}} \frac{d}{d x}(3 x)+\frac{1}{1+(2 x)^{2}} \frac{d}{d x}(2 x) \\ &\frac{d y}{d x}=\frac{1}{1+9 x^{2}}(3)+\frac{1}{1+4 x^{2}}(2) \\ &\frac{d y}{d x}=\frac{3}{1+9 x^{2}}+\frac{2}{1+4 x^{2}} \end{aligned}