Difficulty: Intermediate to Advanced | Prerequisites: Part 1 of these notes (FTC core concepts), chain rule, linearisation
FTC1 in its basic form says d/dx of the integral from a to x of f(t) dt = f(x). But exam problems rarely stay that simple. The upper limit might be x³ instead of x, the lower limit might be a function of x, or both limits might depend on x. All of these require combining FTC1 with the chain rule. This section covers the mechanics of differentiating integrals when the limits are not just plain x and a constant, plus linearisation of accumulation functions.
When the limits of integration are functions of x, apply FTC1 together with the chain rule. Plug the limit function into the integrand and multiply by its derivative. If both limits vary, split the integral at a constant and handle each piece separately.
FTC1 with chain rule (generalised FTC1)
d/dx [integral from a to u(x) of f(t) dt] = f(u(x)) * u'(x). You evaluate the integrand at the upper limit, then multiply by the derivative of that upper limit.
In simple terms, the chain rule "adjusts" for the fact that the upper boundary is moving at a rate different from 1.
Variable lower limit
When the lower limit is g(x) and the upper limit is a constant, flip the integral: integral from g(x) to b of f(t) dt = negative integral from b to g(x) of f(t) dt. Then differentiate the flipped version using the chain rule, which introduces a negative sign.
Both limits variable
When both limits are functions of x, split at any constant c: integral from g(x) to u(x) of f(t) dt = integral from g(x) to c of f(t) dt + integral from c to u(x) of f(t) dt. Differentiate each piece separately.
Linearisation of an accumulation function
The linearisation (tangent-line approximation) of a function f at x = a is L(x) = f(a) + f'(a)(x - a). For an accumulation function defined by an integral, f(a) is the integral evaluated at a (often zero when the limits collapse), and f'(a) comes from FTC1.
d/dx [integral from a to x of f(t) dt] = f(x).
Example: if y = integral from -pi to x of (2 - sin t)/(3 + cos t) dt, then dy/dx = (2 - sin x)/(3 + cos x).
Nothing else to do. Replace the dummy variable t with x.
d/dx [integral from a to u(x) of f(t) dt] = f(u(x)) * u'(x).
Example: if y = integral from sqrt(x) to x³ of sqrt(u) sin(u) du with both limits depending on x, this needs the "both limits" treatment (see below). But if only the upper limit is x³, you get f(x³) 3x².
Flip the integral to make the function-of-x the upper limit. This introduces a negative sign.
d/dx [integral from x to b of f(t) dt] = -f(x).
More generally: d/dx [integral from g(x) to b of f(t) dt] = -f(g(x)) * g'(x).
Example: y = integral from x to 7 of sqrt(2m⁴ + m + 1) dm. Then dy/dx = -sqrt(2x⁴ + x + 1).
Split at any convenient constant c: integral from g(x) to u(x) = integral from g(x) to c + integral from c to u(x).
Differentiate each piece. The lower-limit piece picks up a negative sign.
Result: dy/dx = f(u(x)) u'(x) - f(g(x)) g'(x).
Example: y = integral from x³ to 5 of cos(t)/(t² + 1) dt. Here the upper limit is constant (5) and the lower limit is x³. Flip: y = -integral from 5 to x³ of cos(t)/(t² + 1) dt. So dy/dx = -[cos(x³)/((x³)² + 1)] * 3x² = -3x² cos(x³)/(x⁶ + 1).
Example: y = integral from sqrt(x) to x³ of sqrt(u) sin(u) du. Split at any constant. Then dy/dx = [sqrt(x³) sin(x³)] 3x² - [sqrt(sqrt(x)) sin(sqrt(x))] * (1/(2sqrt(x))).
For f(x) = integral from pi to x of cos³(t) dt, find the linearisation at x = pi.
f(pi) = integral from pi to pi of cos³(t) dt = 0 (limits collapse).
f'(x) = cos³(x) by FTC1, so f'(pi) = cos³(pi) = (-1)³ = -1.
Linearisation: L(x) = 0 + (-1)(x - pi) = -(x - pi) = pi - x.
If F(x) = integral from 1 to x of f(t) dt, then F'(x) = f(x) and F''(x) = f'(x).
When f(t) is itself defined as an integral (e.g. f(t) = integral from 1 to t² of sqrt(1 + u⁴)/u du), you use FTC1 again to find f'(t), then evaluate at the required point.
For instance, F''(2) = f'(2). To find f'(t), differentiate the inner integral: f'(t) = [sqrt(1 + (t²)⁴) / t²] * 2t = 2sqrt(1 + t⁸) / t. Then f'(2) = 2sqrt(1 + 256) / 2 = sqrt(257).
Upper limit u(x): d/dx [integral from a to u(x) of f(t) dt] = f(u(x)) * u'(x)
Lower limit g(x): d/dx [integral from g(x) to b of f(t) dt] = -f(g(x)) * g'(x)
Both limits: d/dx [integral from g(x) to u(x) of f(t) dt] = f(u(x)) u'(x) - f(g(x)) g'(x)
Linearisation: L(x) = f(a) + f'(a)(x - a), where f(a) is the integral at a and f'(a) comes from FTC1
Differentiating integrals with variable limits appears in control theory and signal processing, where a running average or a convolution integral has boundaries that shift with time. In economics, if total cost is accumulated from a variable starting production level to a variable ending level, the marginal effect of shifting either boundary is computed with this technique.
Students forget the chain rule multiplier. If the upper limit is x³, the answer is not just f(x³), it is f(x³) * 3x². The chain rule factor is the single most common error on these problems.
When the lower limit is a function of x and the upper limit is a constant, students often forget the negative sign from flipping the integral. d/dx of integral from x to 7 of f(t) dt = -f(x), not +f(x).
When both limits depend on x, students sometimes only differentiate one of them. You must account for both: the upper-limit term is positive, the lower-limit term is negative.
For linearisation, students sometimes forget that f(a) = 0 when the integral's limits collapse (both limits equal at x = a). The integral from a to a of anything is zero.
⚠️ Differentiating integrals with the chain rule is one of the most frequently tested FTC skills on AP and university calculus exams. Expect at least one problem with a non-trivial upper or lower limit.
⚠️ The "both limits variable" case is a common free-response or short-answer question. Know the split-at-a-constant technique.
⚠️ Linearisation of accumulation functions combines two concepts (FTC1 and tangent-line approximation) and is a favourite "connect the ideas" exam question.
⚠️ Nested integrals (F''(x) where F is an integral of a function that is itself an integral) test whether you can apply FTC1 twice in sequence. These appear as harder exam problems.
Fill in the blank: d/dx [integral from 2 to x⁴ of sin(t) dt] = ______.
sin(x⁴) * 4x³.
True or false: d/dx [integral from x to 5 of e^t dt] = e^x.
False. It equals -e^x (negative sign from the lower limit).
Fill in the blank: If f(x) = integral from pi to x of cos³(t) dt, then f(pi) = ______ and f'(pi) = ______.
f(pi) = 0, f'(pi) = -1.
Q: What is the linearisation of f(x) = integral from pi to x of cos³(t) dt at x = pi?
A: f(pi) = 0 (integral collapses). f'(pi) = cos³(pi) = -1. So L(x) = 0 + (-1)(x - pi) = pi - x. Answer: (E) y = pi - x.
Q: Find dy/dx if y = integral from -pi to x of (2 - sin t)/(3 + cos t) dt.
A: By FTC1, dy/dx = (2 - sin x)/(3 + cos x). Straight substitution of x for t.
Q: Find dy/dx if y = integral from x to 7 of sqrt(2m⁴ + m + 1) dm.
A: Flip limits: y = -integral from 7 to x of sqrt(2m⁴ + m + 1) dm. Then dy/dx = -sqrt(2x⁴ + x + 1).
Q: Find dy/dx if y = integral from x³ to 5 of cos(t)/(t² + 1) dt.
A: Flip: y = -integral from 5 to x³ of cos(t)/(t² + 1) dt. Differentiate: dy/dx = -[cos(x³)/(x⁶ + 1)] * 3x² = -3x² cos(x³)/(x⁶ + 1).
Q: Find dy/dx if y = integral from sqrt(x) to x³ of sqrt(u) * sin(u) du.
A: dy/dx = sqrt(x³) sin(x³) 3x² - sqrt(sqrt(x)) sin(sqrt(x)) 1/(2sqrt(x)) = 3x² x^(3/2) sin(x³) - x^(1/4) * sin(sqrt(x)) / (2sqrt(x)).
Q: If F(x) = integral from 1 to x of f(t) dt, where f(t) = integral from 1 to t² of sqrt(1 + u⁴)/u du, find F''(2).
A: F'(x) = f(x), so F''(x) = f'(x). By the chain rule on f(t): f'(t) = [sqrt(1 + (t²)⁴) / (t²)] * 2t = 2sqrt(1 + t⁸) / t. Then F''(2) = f'(2) = 2sqrt(1 + 256) / 2 = sqrt(257).
This directly extends the chain rule material from earlier differentiation chapters. It also connects to differential equations: the accumulation function g(x) = integral from a to x of f(t) dt is the solution to the initial value problem g' = f, g(a) = 0. Linearisation of accumulation functions ties back to the tangent-line approximation topic and foreshadows Taylor/Maclaurin series, where integrals of known functions are expanded as polynomials.
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