Question

. If f(x) = x|
x| , prove that f'(x) = |2x|.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = x |x|$

Let first define the function, f(x)

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x, \: \: when \: x < 0} \\ &\sf{ \: x, \: \: whenx \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: f(x) = x|x| = \begin{cases} &\sf{ - {x}^{2} , \: \: when \: x < 0} \\ &\sf{ \: {x}^{2} , \: \: whenx \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

Let we first check the differentiability of f(x) at x = 0.

We know,

A function f(x) is said to be differentiable iff

$\boxed{ \rm{ \displaystyle\lim_{x \to a^-} \: \dfrac{f(x) - f(a)}{x - a} = \displaystyle\lim_{x \to a^ + } \: \dfrac{f(x) - f(a)}{x - a}}}$

So, Consider, Left Hand Derivative,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^-} \: \dfrac{f(x) - f(0)}{x - 0}$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^-} \dfrac{ - {x}^{2} - 0}{x }$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^-} \dfrac{ - {x}^{2}}{x }$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^-} ( - x)$

$\rm \:  =  \:  \: 0$

So,

$\bf :\longmapsto\:\displaystyle\lim_{x \to 0^-} \: \dfrac{f(x) - f(0)}{x - 0} = 0$

Now, Consider Right Hand Derivative

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^ + } \: \dfrac{f(x) - f(0)}{x - 0}$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^ + } \dfrac{{x}^{2} - 0}{x }$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^ + } \dfrac{{x}^{2}}{x }$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^ + }x$

$\rm \:  =  \:  \: 0$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^ + } \: \dfrac{f(x) - f(0)}{x - 0} = 0$

Hence, we concluded that,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^-} \dfrac{f(x) - f(0)}{x - 0} = \displaystyle\lim_{x \to 0^ + } \dfrac{f(x) - f(0)}{x - 0}$

$\bf\implies \:f(x) \: is \: differentiable \: at \: x \: = \: 0$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: f'(x) = \begin{cases} &\sf{ - 2{x} , \: \: when \: x < 0} \\ &\sf{ \: 2{x} , \: \: whenx \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \:f'(x) = 2 |x|$