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 :)

## Comments

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TopNewestReally useful, you don't find derivatives of these functions in regular books! – Akshay Yadav · 5 months, 3 weeks ago

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Note: When taking derivatives of continuous and non-differentiable functions, it is important to state the domain of the derivative / the points where the derivative doesn't exist.

This will help prevent mistakes which arise from not realizing that the derivative does not exist, and hence we cannot apply various theorems to it.

E.g. If we want to find the local minimum of \( f(x) = |x| \), we cannot differentiate the function and say "no solutions to \( f'(x) = 0 \) hence no local minimum". – Calvin Lin Staff · 6 months ago

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Really neat work! Loved it. :) – Tapas Mazumdar · 2 months, 1 week ago

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– Sambhrant Sachan · 2 months, 1 week ago

thank youLog in to reply

Yeah, the problem is that in some test questions the domain is not mentioned and I choose "cannot be determined option" but they say that we should stick to the uncommon common sense. -.- – Ashish Siva · 5 months, 3 weeks ago

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Quite useful dude – Ganesh Ayyappan · 6 months ago

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Awesome ! – Vaibhav Prasad · 6 months ago

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– Sambhrant Sachan · 6 months ago

thanks :)Log in to reply

i already knew it ! – Shubham Dhull · 2 months, 2 weeks ago

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That large block of latex full of quads... – Christopher Boo · 5 months ago

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– Upanshu Gupta · 2 weeks, 3 days ago

probably he used an online editor to write the code and pasted it here!Log in to reply