# Differentiate ³√tanx from first principle ?

Jul 30, 2018

$f ' \left(x\right) = {\sec}^{2} \frac{x}{3 {\left(\tan x\right)}^{\frac{2}{3}}}$

#### Explanation:

We know that ,

color(red)((1)(a-b)(a^2+ab+b^2)=a^3-b^3

color(blue)((2)tan(A-B)=(tanA-tanB)/(1+tanAtanB)

color(brown)((3)lim_(theta to0)(tantheta)/(theta)=1

Let , $f \left(x\right) = \sqrt[3]{\tan} x = {\left(\tan x\right)}^{\frac{1}{3}} \implies f \left(t\right) = {\left(\tan t\right)}^{\frac{1}{3}}$

Using First Principle :

$f ' \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x}$

$= {\lim}_{t \to x} \frac{{\left(\tan t\right)}^{\frac{1}{3}} - {\left(\tan x\right)}^{\frac{1}{3}}}{t - x}$

Before applying color(red)((1) Multiply numerator and denominator by

$\left[{\left(\tan t\right)}^{\frac{2}{3}} + {\left(\tan t\right)}^{\frac{1}{3}} {\left(\tan x\right)}^{\frac{1}{3}} + {\left(\tan x\right)}^{\frac{2}{3}}\right]$

$f ' \left(x\right) = {\lim}_{t \to x}${color(red)(((tant)^(1/3)-(tanx)^(1/3)))/(t-x) [color(red)((tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3))}/[(tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3]}

Apply color(red)((1)  to numerator

=lim_(t tox)color(red)((tant-tanx))/(t-x)1/[(tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3)

$= \frac{1}{{\left(\tan x\right)}^{\frac{2}{3}} + {\left(\tan x\right)}^{\frac{1}{3}} {\left(\tan x\right)}^{\frac{1}{3}} + {\left(\tan x\right)}^{\frac{2}{3}}} {\lim}_{t \to x} \frac{\tan t - \tan x}{t - x}$
$= \frac{1}{3 {\left(\tan x\right)}^{\frac{2}{3}}} \times {\lim}_{t \to x} \frac{\tan t - \tan x}{t - x}$

Before applying color(blue)((2) Multiply numerator and denominator by

$\left(1 + \tan t \tan x\right)$

$f ' \left(x\right) = \frac{1}{3 {\left(\tan x\right)}^{\frac{2}{3}}} {\lim}_{t \to x} \frac{\textcolor{b l u e}{\left(\tan t - \tan x\right)}}{t - x} \cdot \frac{1 + \tan t \tan x}{\textcolor{b l u e}{\left(1 + \tan t \tan x\right)}}$

$\therefore f ' \left(x\right) = \frac{1}{3 {\left(\tan x\right)}^{\frac{2}{3}}} {\lim}_{t \to x} \frac{\textcolor{b l u e}{\frac{\tan t - \tan x}{1 + \tan t \tan x}}}{t - x} \cdot {\lim}_{t \to x} \left(1 + \tan t \tan x\right)$

$\therefore f ' \left(x\right) = \frac{1}{3 {\left(\tan x\right)}^{\frac{2}{3}}} \cdot \left(1 + \tan x \tan x\right) {\lim}_{t \to x} \frac{\textcolor{b l u e}{\frac{\tan t - \tan x}{1 + \tan t \tan x}}}{t - x}$

Using color(blue)((2) we get

$f ' \left(x\right) = \frac{1}{3 {\left(\tan x\right)}^{\frac{2}{3}}} \left(1 + {\tan}^{2} x\right) \textcolor{b r o w n}{_ \left(\left(t - x\right) \to 0\right) \left[\tan \frac{t - x}{t - x}\right]}$

:.f'(x)=sec^2x/(3(tanx)^(2/3) ) xxcolor(brown)((1))...tocolor(brown)(Apply(3)

$\implies f ' \left(x\right) = {\sec}^{2} \frac{x}{3 {\left(\tan x\right)}^{\frac{2}{3}}}$

Jul 30, 2018

${\sec}^{2} \frac{x}{3 {\tan}^{\frac{2}{3}} x} .$

#### Explanation:

We know that, $f ' \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x}$.

So, if $f \left(x\right) = \sqrt[3]{\tan} x = {\tan}^{\frac{1}{3}} x$, then,

$f ' \left(x\right) = {\lim}_{t \to x} \frac{{\tan}^{\frac{1}{3}} t - {\tan}^{\frac{1}{3}} x}{t - x}$,

$= \lim \frac{{\tan}^{\frac{1}{3}} t - {\tan}^{\frac{1}{3}} x}{\tan t - \tan x} \cdot \frac{\tan t - \tan x}{t - x}$,

$i . e . , f ' \left(x\right) = {L}_{1} \cdot {L}_{2} \text{............................."(ast)," say, where, }$

${L}_{1} = {\lim}_{t \to x} \frac{{\tan}^{\frac{1}{3}} t - {\tan}^{\frac{1}{3}} x}{\tan t - \tan x}$.

Here, let ${\tan}^{\frac{1}{3}} t = T \mathmr{and} {\tan}^{\frac{1}{3}} x = X$.

:." As "t to x, T to X; and, tant=T^3, tanx=X^3.

$\therefore {L}_{1} = {\lim}_{T \to X} \frac{T - X}{{T}^{3} - {X}^{3}}$,

$= \lim \frac{T - X}{\left(T - X\right) \left({T}^{2} + T X + {X}^{2}\right)}$,

$= {\lim}_{T \to X} \frac{1}{{T}^{2} + T X + {X}^{2}}$,

$= \frac{1}{{X}^{2} + X \cdot X + {X}^{2}}$.

$\Rightarrow {L}_{1} = \frac{1}{3 {X}^{2}} = \frac{1}{3 {\tan}^{\frac{2}{3}} x} \ldots \ldots \ldots \ldots \ldots . . \left({\ast}^{1}\right)$.

${L}_{2} = {\lim}_{t \to x} \frac{\tan t - \tan x}{t - x}$,

$= \lim \frac{\sin \frac{t}{\cos} t - \sin \frac{x}{\cos} x}{t - x}$,

$= \lim \frac{\sin t \cos x - \cos t \sin x}{\left(\cos t\right) \left(\cos x\right) \left(t - x\right)}$,

$= {\lim}_{t \to x} \left\{\frac{\sin \left(t - x\right)}{t - x}\right\} \sec t \cdot \sec x$,

$= 1 \cdot \sec x \cdot \sec x$.

$\Rightarrow {L}_{2} = {\sec}^{2} x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({\ast}^{2}\right)$.

Combining $\left(\ast\right) , \left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right)$, we have,

$\left[\sqrt[3]{\tan} x\right] ' = {\sec}^{2} \frac{x}{3 {\tan}^{\frac{2}{3}} x}$, as Respected Maganbhai P.

has derived!