Difficulty: Intermediate | Prerequisites: Calc I (basic derivatives, chain rule, FTC introduction)
The Fundamental Theorem of Calculus (FTC) connects differentiation and integration. In Calc II, you regularly face functions defined as integrals with variable limits, and the core task is to differentiate them. This builds directly on the chain rule from Calc I, so if your chain rule is rusty, shore that up first. The skill shows up in nearly every exam and is the gateway to later topics like improper integrals and series.
When a function is defined as a definite integral whose limits depend on x, you differentiate it by plugging the limit into the integrand and multiplying by that limit's derivative. If both limits are variable, split it into two pieces using the additive property of integrals and differentiate each.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b] and F(x) = the integral from a to x of f(t) dt, then F'(x) = f(x). In simple terms, differentiating an integral "undoes" the integration, and you just get back the integrand evaluated at the upper limit.
Variable limit of integration
An upper or lower bound of a definite integral that is itself a function of the differentiation variable (e.g. sin(6x) or sqrt(t) rather than a constant). Think of it as the boundary of the integral moving as x changes.
Chain rule (in the FTC context)
When the upper limit is not simply x but a function u(x), you must multiply the integrand evaluated at u(x) by u'(x). In simple terms, the moving boundary's own rate of change scales the result.
Integrand
The function being integrated, written between the integral sign and the differential (e.g. ln(1 + t^4) in the integral of ln(1 + t^4) dt). Think of it as the "inside function" that FTC plugs the limit into.
This is the most common setup. If f(x) = integral from a to u(x) of g(t) dt, then:
f'(x) = g(u(x)) multiplied by u'(x)
The constant lower limit contributes nothing (its derivative is zero)
Worked Example (PS1, 1a)
f(x) = integral from -3pi to sin(6x) of ln(1 + t^4) dt
The integrand is g(t) = ln(1 + t^4), which is continuous on all of R
The upper limit is u(x) = sin(6x), so u'(x) = 6 cos(6x)
The lower limit is the constant -3pi, so its contribution is zero
Apply FTC1 with chain rule: f'(x) = ln(1 + sin^4(6x)) multiplied by 6 cos(6x)
Result: f'(x) = 6 cos(6x) ln(1 + sin^4(6x))
When both limits depend on the variable, split the integral at any convenient constant c using the additive property:
Integral from a(t) to b(t) of g(theta) d(theta) = integral from c to b(t) minus integral from c to a(t)
Then differentiate each piece separately.
Worked Example (PS1, 1b)
h(t) = integral from t to sqrt(t) of (sin(theta^8))^2026 d(theta)
The integrand is g(theta) = (sin(theta^8))^2026, continuous on R
Split at any constant (say 0): h(t) = integral from 0 to sqrt(t) minus integral from 0 to t
Upper piece: upper limit is sqrt(t) = t^(1/2), derivative is (1/2) t^(-1/2)
Contribution: (sin((sqrt(t))^8))^2026 multiplied by (1/2) t^(-1/2) = (sin(t^4))^2026 / (2 sqrt(t))
Lower piece: upper limit is t, derivative is 1
Contribution: (sin(t^8))^2026 multiplied by 1
Subtract: h'(t) = (sin(t^4))^2026 / (2 sqrt(t)) minus (sin(t^8))^2026
Result: h'(t) = (1 / (2 sqrt(t))) (sin(t^4))^2026 - (sin(t^8))^2026
For F(x) = integral from a(x) to b(x) of g(t) dt:
F'(x) = g(b(x)) multiplied by b'(x) minus g(a(x)) multiplied by a'(x)
The upper limit gets a positive sign, the lower limit gets a negative sign
Always check continuity of the integrand before applying FTC1
\frac{d}{dx}\int_{a}^{u(x)} g(t)\,dt = g\bigl(u(x)\bigr)\cdot u'(x)\frac{d}{dx}\int_{a(x)}^{b(x)} g(t)\,dt = g\bigl(b(x)\bigr)\cdot b'(x) \;-\; g\bigl(a(x)\bigr)\cdot a'(x)The first formula covers the standard case (constant lower limit). The second covers the general case where both limits are functions of x.
Students often forget to multiply by the derivative of the upper limit. FTC1 alone gives you g(u(x)), but the chain rule factor u'(x) is essential. Without it, your answer is incomplete.
Students sometimes think you need to find an antiderivative of the integrand first. You do not. FTC1 lets you differentiate directly without ever integrating.
When the lower limit is a function (not a constant), students often give it the wrong sign. The lower-limit term is subtracted, not added.
Students occasionally plug the limits back in as though they are evaluating a definite integral. Differentiating an integral-defined function is a different operation from computing the integral's value.
FTC differentiation appears on virtually every Calc II exam, often as the first problem. It is meant to be quick marks if you know the formula.
Expect at least one problem where both limits are variable. The split-at-a-constant technique is the standard approach.
Professors like to pair tricky inner functions (trig compositions, nested powers) with FTC to test whether you can handle the chain rule under pressure.
Always state that the integrand is continuous before applying FTC1. Some graders dock marks if you skip this check.
True or False: To differentiate an integral from a to u(x) of g(t) dt, you first need to find the antiderivative of g. (False)
Fill in the blank: When using FTC1 with a variable upper limit u(x), you multiply the integrand evaluated at ____ by ____. (u(x); u'(x))
True or False: If the lower limit is a constant, it contributes a term to the derivative. (False)
Fill in the blank: When both limits are variable, split the integral at a ____ to create two single-variable-limit integrals. (constant)
True or False: The lower-limit contribution in the general formula carries a negative sign. (True)
Q: Find d/dx of the integral from 2 to x^3 of e^(t^2) dt.
A: By FTC1 with chain rule, the answer is e^((x^3)^2) times 3x^2 = 3x^2 e^(x^6).
Q: Find d/dt of the integral from t^2 to cos(t) of sqrt(1 + u^3) du.
A: Split at a constant. Upper-limit contribution: sqrt(1 + cos^3(t)) times (-sin(t)). Lower-limit contribution (subtracted): sqrt(1 + t^6) times 2t. Final answer: -sin(t) sqrt(1 + cos^3(t)) - 2t sqrt(1 + t^6).
Q: Why must you verify that the integrand is continuous before applying FTC1?
A: FTC1 requires that the integrand be continuous on the interval of integration. If it has a discontinuity, the theorem does not apply and the derivative formula is not valid.
Q: A student writes d/dx of the integral from 0 to sin(x) of t^3 dt = sin^3(x). What did they get wrong?
A: They forgot the chain rule. The correct answer is sin^3(x) times cos(x), because the derivative of the upper limit sin(x) is cos(x).
This connects directly to accumulation functions (Problem 2 in this set), where g(x) = integral from a to x of f(t) dt defines a new function whose derivative is f(x). That relationship is the engine behind analysing increasing/decreasing behaviour and finding extrema of accumulation functions.
FTC differentiation also surfaces later in the course when working with improper integrals and when setting up integrals in applications (arc length, surface area), where you differentiate under the integral sign.
FTC, Fundamental Theorem of Calculus Part 1, Leibniz integral rule, differentiating under the integral sign, variable limits of integration, chain rule with integrals, FTC with substitution, Calc 2 NYU, integral-defined functions, accumulation function derivative