Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation  Exercise 10.3 Question 30 Maths Textbook Solution.

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

Hint:

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

Given:

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

Solution:

Let, $y=\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)$

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

Since,$3x-2x=x$

\begin{aligned} &y=\tan ^{-1} 3 x-\tan ^{-1} 2 x \\ &{\left[\text { Since }+\tan x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]} \end{aligned}

Differentiating it with respect tox using chain rule,

\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+\left(3 x^{2}\right)} \frac{d}{d x}(3 x)-\frac{1}{1+(2 x)^{2}} \frac{d}{d x}(2 x) \\ &\left\{\text { Since } \Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\tan ^{-1}(x)=\frac{1}{1+x^{2}}\right\} \\ &\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}