Question

Find the derivative of modulus function,
$f(x) = |x|$

Answer is given as x/|x|, I need explanation.​

Answer 1

ANSWER:

To Differentiate:

• f(x) = |x|

Solution:

We are given that,

$\implies f(x) =|x|$

We know that,

$\implies |x|=\sqrt{x^2}$

So,

$\implies f(x)=\sqrt{x^2}$

We can re-write it as,

$\implies f(x)=(x^2)^{\frac{1}{2}}$

Differentiating both sides w.r.t. x,

$\implies\dfrac{d}{dx}f(x)=\dfrac{d}{dx}(x^2)^{\frac{1}{2}}$

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}(x^2)^{\frac{1}{2}}$

We know that,

$\hookrightarrow\dfrac{d}{dx}x^n=nx^{n-1}$

So,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}(x^2)^{\frac{1}{2}}$

$\implies\dfrac{dy}{dx}=\dfrac{1}{2}(x^2)^{\frac{1}{2}-1}\times\dfrac{d}{dx}(x^2)$

$\implies\dfrac{dy}{dx}=\dfrac{1}{2}(x^2)^{\frac{-1}{2}}\times2(x)^{2-1}$

$\implies\dfrac{dy}{dx}=\dfrac{1}{2}\cdot\dfrac{1}{(x^2)^{\frac{1}{2}}}\times2x$

$\implies\dfrac{dy}{dx}=\dfrac{1}{2\!\!\!/}\cdot\dfrac{1}{\sqrt{(x^2)}}\times2\!\!\!/\:x$

$\implies\dfrac{dy}{dx}=\dfrac{x}{\sqrt{(x^2)}}$

As,

$\implies |x|=\sqrt{x^2}$

So,

$\implies\dfrac{dy}{dx}=\dfrac{x}{\sqrt{x^2}}$

$\implies\dfrac{dy}{dx}=\dfrac{x}{|x|}$

Hence,

$\implies\dfrac{d}{dx}f(x)=\dfrac{x}{|x|}$

Therefore,

$\bf\implies\dfrac{d}{dx}|x|=\dfrac{x}{|x|}$