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 $x^2$.

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