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 $x^2$.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.