Question

Differentiate the function

$|x| \: with \: respect \: to \: x \: provided \: that \: x \ne \: 0$
​

Answer 1

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

Given function is

$\rm :\longmapsto\: |x|$

Let assume that

$\rm :\longmapsto\: y = |x|$

can be rewritten as

$\rm :\longmapsto\: y = \sqrt{ {x}^{2} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}y = \dfrac{d}{dx} \sqrt{ {x}^{2} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{2} } }\dfrac{d}{dx} {x}^{2}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: \: \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{2} } } \times {2x}^{2 - 1}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{x}{ |x| } \: \: \green{and \: x \: \ne \: 0}$

Hence,

$\bf\implies \:\boxed{\tt{ \dfrac{d}{dx} |x| = \dfrac{x}{ |x| } \: \: \green{and \: x \: \ne \: 0}}}$

OR

$\purple{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\dfrac{d}{dx} |x| = \begin{cases} &\sf{ - 1 \: \: \: when \: x \: < \: 0} \\ \\ &\sf{ \: \: 1 \: \: \: \: when \: x \: > \: 0} \end{cases}\end{gathered}\end{gathered}}$

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More to Know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

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

Given function is

$\rm :\longmapsto\: |x|$

Let assume that

$\rm :\longmapsto\: y = |x|$

can be rewritten as

$\rm :\longmapsto\: y = \sqrt{ {x}^{2} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}y = \dfrac{d}{dx} \sqrt{ {x}^{2} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{2} } }\dfrac{d}{dx} {x}^{2}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: \: \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{2} } } \times {2x}^{2 - 1}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{x}{ |x| } \: \: \green{and \: x \: \ne \: 0}$

Hence,

$\bf\implies \:\boxed{\tt{ \dfrac{d}{dx} |x| = \dfrac{x}{ |x| } \: \: \green{and \: x \: \ne \: 0}}}$

OR

$\purple{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\dfrac{d}{dx} |x| = \begin{cases} &\sf{ - 1 \: \: \: when \: x \: < \: 0} \\ \\ &\sf{ \: \: 1 \: \: \: \: when \: x \: > \: 0} \end{cases}\end{gathered}\end{gathered}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$