Math Doubts


A differentiable function whose derivative is equal to the original function, is called an antiderivative of a function. The antiderivative of a function is also called as a primitive of a function.


$f{(x)}$ is a function and $p{(x)}$ is a differentiable function. Assume, the derivative of a function $p{(x)}$ is equal to $f{(x)}$ and $C$ is a constant.

$\dfrac{d}{dx}{\Big(p{(x)}+C\Big)}$ $\,=\,$ $f{(x)}$

$\implies$ $p'{(x)} \,=\, f{(x)}$

The function $p{(x)}$ is called an antiderivative of function $f{(x)}$.


$\dfrac{x^3}{3}$ is a function and $x^2$ is another function. Now, differentiate the function $\dfrac{x^3}{3}$ with a constant.

$\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $\dfrac{d}{dx}{\dfrac{x^3}{3}}+\dfrac{d}{dx}{\,C}$

$\implies$ $\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $\dfrac{1}{3} \times \dfrac{d}{dx}{x^3}+0$

$\implies$ $\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $\dfrac{1}{3} \times 3x^{3-1}$

$\implies$ $\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $\dfrac{1}{3} \times 3x^2$

$\implies$ $\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $\dfrac{3x^2}{3}$

$\implies$ $\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{3}x^2}{\cancel{3}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\Big(\dfrac{x^3}{3}+C\Big)}$ $\,=\,$ $x^2$

Therefore, the function $\dfrac{x^3}{3}$ is called a primitive or antiderivative of the function $x^2$.

Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved