Derivative of peculiar functions

Differentiation is considered an easy topic in calculus. Far more easier than integration, but Some functions are just made to be painful. Today i will be discussing with you guys the derivative of these functions . There are mainly 3 of them \(|x| ,\lfloor x\rfloor\) and \(\{ x\} \) . Let's Begin !

$1) \quad |x|$

There are two ways to calculate the derivative of $|x|$ . If you know any other method , please share it in the comments section below.

$\text{Let } v=|x| \implies v^2=x^2 \quad \quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\text{: Alternate :} \\ \text{Take Derivative w.r.t to x on both sides} \quad \quad\quad\quad\quad\quad\quad \quad\quad\quad\quad f(x)=|x|=\sqrt{x^2} \\ 2v\dfrac{dv}{dx}=2x \implies \dfrac{dv}{dx}=\dfrac{x}{v} \quad \quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad f^{'}(x)=\dfrac{1}{2\sqrt{x^2}}\cdot2x \\ \boxed{\dfrac{dv}{dx}=\dfrac{x}{|x|}} \quad \quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \boxed{f^{'}(x)=\dfrac{x}{|x|}}$

The Result makes a lot of sense .... if $x>0$, the Derivative is $1$ and if $x<0$, Derivative is $-1$

This is indeed what the graph of $|x|$ tells us!

Now Let's take a general Case of $f(x)=|ax^b+c|^d$, Where $a,b,c,d \in \mathbb R$

$f^{'}(x)=d\cdot|ax^b+c|^{d-1}\cdot\dfrac{d\left(|ax^b+c|\right)}{dx} \\ f^{'}(x)=d\cdot|ax^b+c|^{d-1}\cdot\dfrac{ax^b+c}{|ax^b+c|}\cdot ab\cdot x^{b-1} \\ \boxed{f^{'}(x)=abd\cdot x^{b-1}\cdot|ax^b+c|^{d-2}\cdot{(ax^b+c)}}$

$2) \quad \lfloor x\rfloor$

When floor function or Greatest integer function is applied on a function , the range changes to integral values. We can say that the derivative of the function is $0$ when the function is continuous. The derivative does not exists at integer values of x, because the graph is discontinuous .

For a General $f(x)= \lfloor ax^b+c \rfloor^{d}$, the derivative will always be $\boxed{0}$ .

$3) \quad \{ x \}$

When Fractional part function is applied on a function , the Range changes to $[0,1)$ .

$\{ f(x) \}$ can be written as $f(x)-\lfloor f(x)\rfloor$. Using this property we will find the derivative of $\{ x \}$ . The derivative does not exists at integer values of $x$ , because the graph is discontinuous .

Let $f(x) = \{ x \} = x - \lfloor x \rfloor \\ f^{'}(x)=1 - 0 =1$

For a General $f(x) = \{(ax^b+c)^d\} = (ax^b+c)^d - \lfloor (ax^b+c)^d \rfloor$

$f^{'}(x) = d\cdot(ax^b+c)^{d-1}\cdot abx^{b-1} - 0 \\ \boxed{f^{'}(x) = abd\cdot x^{b-1}(ax^b+c)^{d-1}}$

I Hope you enjoyed this :)