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Differentiability

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Question

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

Moderate

Solution

$g^{1} (0) = b = \underset{x \to 0}{L t} \frac{x^{2} + x \tan x - x \tan 2 x}{x (a x + \tan x - \tan 3 x)}$
$\begin{array}{l} = \underset{x \to 0}{L t} \frac{x + \tan x - \tan 2 x}{a x + \tan x - \tan 3 x} \\ = \underset{x \to 0}{L t} \frac{x (x + \frac{x^{3}}{3} + \frac{2}{15} x^{5} + ------- \infty) - (2 x + \frac{8 x^{3}}{3} + ----- \infty)}{a x (x + \frac{x^{3}}{3} + \frac{2}{15} x^{5} + ------- \infty) - (3 x + \frac{27 x^{3}}{3} + ----- \infty)} \end{array}$
$On simplifying a = \underline{\underline{2}}, b = \frac{7}{26}$

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