$\text{ Let }g\left(x\right)=\left\{\begin{array}{cc}\frac{x^2+x\tan x-x\tan 2x}{ax+\tan x-\tan 3x},& x\ne 0\\ 0& ,x=0\end{array}\text{ If }{g}^1\left(0\right)\right.\text{ exists and is equal to non-zero value b then }$$\text{b is equal to }$

$g^1\left(0\right)=b=\underset{x\to 0}{Lt}\frac{x^2+x\tan x-x\tan 2x}{x\left(ax+\tan x-\tan 3x\right)}$

$\begin{array}{l}=\underset{x\to 0}{Lt}\frac{x+\tan x-\tan 2x}{ax+\tan x-\tan 3x}\\ =\underset{x\to 0}{Lt}\frac{x\left(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+\text{-------}\infty \right)-\left(2x+\frac{8x^3}{3}+\text{-----}\infty \right)}{ax\left(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+\text{-------}\infty \right)-\left(3x+\frac{27x^3}{3}+\text{-----}\infty \right)}\end{array}$

$\text{ On simplifying }\mathrm{a}=\underset{\_}{\underset{\_}{2}},\mathrm{b}=\frac{7}{26}$