Why Is the Derivative of an Antiderivative Equal to the Original Function?

The statement that the derivative of an antiderivative equals the original function has two closely related meanings. If $F$ is already specified to be an antiderivative of $f$, the statement is the definition of “antiderivative.” The Fundamental Theorem of Calculus gives a substantive result: it shows that integrating a continuous function from a fixed starting point produces a function whose derivative is the original function.

What “antiderivative” means

A function $F$ is an antiderivative, or primitive, of $f$ on an interval if

$$F'(x)=f(x)$$

at every point in that interval where the derivative is considered. So there is no separate proof that an antiderivative has derivative $f$; that property is what makes it an antiderivative.

For instance, $F(x)=x^3/3$ is an antiderivative of $f(x)=x^2$ because differentiating gives $F'(x)=x^2$. Every function $x^3/3+C$, where $C$ is constant, is also an antiderivative, since the derivative of a constant is zero.

The deeper question is why an integral-defined function has this property.

Fundamental Theorem of Calculus: derivative of an integral

Suppose $f$ is continuous on an interval $I$ containing $a$, and define

$$F(x)=\int_a^x f(t)\,dt$$

for $x\in I$. Then the first part of the Fundamental Theorem of Calculus states that $F$ is differentiable at every interior point $x$ of $I$ and

$$F'(x)=f(x).$$

At an endpoint of $I$, the corresponding one-sided derivative satisfies the same formula. The continuity assumption is a convenient standard condition that guarantees the result at every interior point of the interval. It can be weakened, but continuity is the usual introductory-calculus version.

Proof from the difference quotient

Fix an interior point $x$ and take a nonzero increment $h$ small enough that $x+h$ is in the interval. By the definition of $F$,

$$\begin{aligned} \frac{F(x+h)-F(x)}{h} &=\frac{1}{h}\left(\int_a^{x+h} f(t)\,dt-\int_a^x f(t)\,dt\right)\\ &=\frac{1}{h}\int_x^{x+h} f(t)\,dt. \end{aligned}$$

This quotient is the average value of $f$ over the short interval from $x$ to $x+h$ (with the usual signed-integral convention if $h<0$). As $h$ approaches zero, continuity means that values of $f(t)$ on this interval get arbitrarily close to $f(x)$. Therefore their average gets close to $f(x)$ as well. In the limit,

$$F'(x)=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=f(x).$$

A precise estimate makes the argument explicit. Subtract $f(x)$ from the quotient:

$$\left|\frac{1}{h}\int_x^{x+h}f(t)\,dt-f(x)\right| \leq \sup_{t\text{ between }x\text{ and }x+h}|f(t)-f(x)|.$$

By continuity at $x$, the right-hand side tends to zero as $h\to0$. That proves the derivative formula. At an endpoint of $I$, the same estimate applies with $h$ restricted to increments that stay in $I$, proving the corresponding one-sided derivative formula.

Worked example

Let $f(t)=t^2$ and define $F(x)=\int_1^x t^2\,dt$. Evaluating the integral gives

$$F(x)=\frac{x^3-1}{3}.$$

Differentiate this expression:

$$F'(x)=\frac{1}{3}(3x^2)=x^2=f(x).$$

The theorem also gives the derivative directly, without first evaluating the integral: since the integrand $t^2$ is continuous, $F'(x)=x^2$.

Common mistakes and variations

• Confusing the definition with the theorem: “$F$ is an antiderivative of $f$” already asserts $F'=f$. The theorem explains why a particular integral-defined function is an antiderivative.
• Dropping the continuity condition: For a continuous $f$, the derivative formula holds throughout the interval’s interior, with the corresponding one-sided formula at endpoints. With less regular functions, the integral-defined function need not have the stated derivative at every point.
• Mixing up the variable of integration: In $\int_a^x f(t)\,dt$, $t$ is a dummy variable and $x$ is the endpoint. The result is a function of $x$.
• Forgetting the constant: If $F'=f$, then $F+C$ has the same derivative. An indefinite integral, $\int f(x)\,dx$, represents a family of antiderivatives, not just one function.

A related part of the Fundamental Theorem connects definite integrals to any antiderivative: if $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'(x)=f(x)$ there, with $f$ Riemann integrable on $[a,b]$, then $\int_a^b f(x)\,dx=F(b)-F(a)$. This is often called the evaluation part of the theorem.