The derivative of an integral with x in the limits

By Marius, private math tutor in Montreal · 2026-10-05

To differentiate $\int_{a}^{x} f (t) d t$ when $f$ is continuous, put $x$ in place of $t$ in the integrand: the derivative is $f (x)$ . If the upper limit is a function $g (x)$ , multiply by $g^{'} (x)$ , and if both limits depend on $x$ , subtract the lower limit's term from the upper limit's: $f (g (x)) g^{'} (x) - f (h (x)) h^{'} (x)$ .

Part 1 of the Fundamental Theorem

If $f$ is continuous on an open interval that contains $a$ , then the accumulation function

F (x) = \int_{a}^{x} f (t) d t

is differentiable on that interval, and

F^{'} (x) = f (x)

$t$ is a dummy variable. The integration uses it up; the answer is a function of $x$ alone.
The constant $a$ does not appear in the answer. Moving the lower limit adds a constant to $F$ , and a constant differentiates to zero.
No antiderivative is needed. The theorem works just as well for $\frac{\sin t}{t}$ , which has no antiderivative built from the usual functions, as it does for $t^{2}$ .

Why it is true

Moving the upper limit from $x$ to $x + h$ adds one thin strip of area: $F (x + h) - F (x) = \int_{x}^{x + h} f (t) d t$
By the Mean Value Theorem for Integrals, that strip equals a rectangle whose height is a value of $f$ inside it: for some $c_{h}$ between $x$ and $x + h$ , $\int_{x}^{x + h} f (t) d t = f (c_{h}) h$
Divide by $h$ : $\frac{F (x + h) - F (x)}{h} = f (c_{h})$ .
As $h \to 0$ , $c_{h}$ is squeezed onto $x$ . Since $f$ is continuous, $f (c_{h}) \to f (x)$ , and that limit is $F^{'} (x)$ .

In words: the area grows at a rate equal to the height of the graph at the moving edge.

When the upper limit is a function of x

If $G (x) = \int_{a}^{g (x)} f (t) d t$ , then $G$ is the accumulation function $F$ above, evaluated at $g (x)$ : $G (x) = F (g (x))$ . The chain rule gives

G^{'} (x) = F^{'} (g (x)) g^{'} (x) = f (g (x)) g^{'} (x)

Substitute the upper limit into the integrand, then multiply by the derivative of the upper limit.

Worked example

Find $G^{'} (x)$ for $G (x) = \int_{1}^{\sqrt{x}} \frac{\sin t}{t} d t$ , where $x > 0$ .

The upper limit is $g (x) = \sqrt{x}$ , so $g^{'} (x) = \frac{1}{2 \sqrt{x}}$ .
Substitute it into the integrand: $f (\sqrt{x}) = \frac{\sin \sqrt{x}}{\sqrt{x}}$ .
Multiply: $G^{'} (x) = \frac{\sin \sqrt{x}}{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}} = \frac{\sin \sqrt{x}}{2 x}$

No antiderivative of $\frac{\sin t}{t}$ was needed, which is fortunate, since none can be written in terms of the usual functions.

When both limits depend on x

Split the integral at any constant $c$ in an interval where $f$ is continuous and that contains both limits, so that each piece has a constant lower limit:

\int_{h (x)}^{g (x)} f (t) d t = \int_{c}^{g (x)} f (t) d t - \int_{c}^{h (x)} f (t) d t

Each piece is the case above, so

\frac{d}{d x} \int_{h (x)}^{g (x)} f (t) d t = f (g (x)) g^{'} (x) - f (h (x)) h^{'} (x)

The constant $c$ has vanished, so you never have to choose one. The split only shows where the formula comes from.
A constant limit has derivative 0, so its term drops out. With only the lower limit moving, the derivative is $- f (h (x)) h^{'} (x)$ : the minus sign belongs to the lower limit.
The formula needs $f$ continuous on an interval containing every value the two limits take. If the integrand blows up between the limits, the integral is improper and this is not the tool.

Worked example

Find $H^{'} (x)$ for $H (x) = \int_{x^{2}}^{3 x} \cos (t^{2}) d t$ , and its value at $x = 0$ .

Upper limit: $g (x) = 3 x$ , so $g^{'} (x) = 3$ , and the integrand there is $\cos (9 x^{2})$ .
Lower limit: $h (x) = x^{2}$ , so $h^{'} (x) = 2 x$ , and the integrand there is $\cos (x^{4})$ .
Upper term minus lower term: $H^{'} (x) = 3 \cos (9 x^{2}) - 2 x \cos (x^{4})$
At $x = 0$ : $H^{'} (0) = 3 \cos 0 - 0 = 3$ .

Each limit brings its own derivative as a factor, and the lower limit's term is subtracted. Two slips are the usual ones. One is substituting the limits into the integrand and stopping, which loses $g^{'} (x)$ and $h^{'} (x)$ . The other is adding the lower limit's term when it should be subtracted.

When x is inside the integrand as well

Everything above assumes the integrand contains $t$ only. If $x$ appears inside too, move it out first. While $t$ runs between the limits, $x$ stays fixed, so for the integral it is a constant:

K (x) = \int_{1}^{x} x e^{t^{2}} d t = x \int_{1}^{x} e^{t^{2}} d t

That is a product, and the product rule comes before Part 1:

K^{'} (x) = \int_{1}^{x} e^{t^{2}} d t + x e^{x^{2}}

The first term stays an integral. Nothing asks for it to be evaluated, and it could not be in elementary terms anyway.

Check your understanding

No calculator needed.

Find $\frac{d}{d x} \int_{2}^{x} \ln (1 + t^{2}) d t$ .
Find $\frac{d}{d x} \int_{0}^{5 x} e^{\sin t} d t$ .
True or false: if $a$ and $b$ are constants, the derivative of $\int_{a}^{b} f (t) d t$ with respect to $x$ is $f (b) - f (a)$ . Explain.
Find $\frac{d}{d x} \int_{e^{x}}^{5} \ln t d t$ .
Let $H (x) = \int_{x}^{x^{3}} \cos (π t) d t$ . Find $H^{'} (1)$ .
To differentiate $\int_{x^{2}}^{\sin x} f (t) d t$ , where $f$ is continuous everywhere, one student splits the integral at $t = 0$ and another at $t = 7$ . Which of them gets the right derivative, and why?
Let $G (x) = \int_{0}^{x} x \cos t d t$ . Find $G^{'} (x)$ , and explain why $x \cos x$ is not the answer.

Show the answers

$\ln (1 + x^{2})$ . The integrand is continuous everywhere and the upper limit is plain $x$ , so Part 1 applies directly: replace $t$ by $x$ .
$5 e^{\sin 5 x}$ . Substitute $5 x$ for $t$ , then multiply by the derivative of the upper limit, which is 5.
False: the derivative is 0. With two constant limits the integral is a single number, and a number does not change with $x$ . The expression $f (b) - f (a)$ mixes up the two parts of the theorem: Part 2 evaluates the integral as $F (b) - F (a)$ , with an antiderivative $F$ , not with $f$ .
$- x e^{x}$ . Only the lower limit moves, so the derivative is $- \ln (e^{x}) \cdot e^{x} = - x e^{x}$ . As a check, $t \ln t - t$ is an antiderivative of $\ln t$ , and differentiating ${[t \ln t - t]}_{e^{x}}^{5}$ gives the same.
$- 2$ . $H^{'} (x) = \cos (π x^{3}) \cdot 3 x^{2} - \cos (π x) \cdot 1$ , so $H^{'} (1) = 3 \cos π - \cos π = - 3 + 1 = - 2$ .
Both, and they get the same answer, $f (\sin x) \cos x - 2 x f (x^{2})$ . Moving the split point from 0 to 7 changes each piece by the same constant, $\int_{0}^{7} f (t) d t$ , and a constant differentiates to 0.
$G^{'} (x) = \sin x + x \cos x$ . Since $x$ is constant while $t$ varies, $G (x) = x \int_{0}^{x} \cos t d t = x \sin x$ , and the product rule gives two terms. The answer $x \cos x$ differentiates only the integral, treating the factor $x$ in front as if it did not change.

Practise it

The free CEGEP Calculus II worksheet on the definite integral and the Fundamental Theorem has a sheet of Part 1 derivatives, including an upper limit of $x^{3}$ and $x$ as the lower limit, alongside Part 2, Riemann sums and average value. For the chain rule itself, the Calculus I differentiation rules worksheet is the place to start.

Want one-on-one help with this? I offer CEGEP Science calculus and linear algebra tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.

The derivative of an integral with x in the limits

Part 1 of the Fundamental Theorem

Why it is true

When the upper limit is a function of x

Worked example

When both limits depend on x

Worked example

When x is inside the integrand as well

Check your understanding

Practise it

Practise with a free worksheet

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