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A JEE question gives you F(x) = ∫ from x to x² of (sin t)/t dt and asks for F′(x). You cannot write that integral in closed form, and you do not need to. One short rule, usually called the Leibniz rule, turns these questions into two lines of work. This post shows the rule, why it is true, five worked examples and the mistakes examiners count on.
Where is this in the syllabus?
The NCERT Class 12 chapter on Integrals (2026-27 reprint) states the first fundamental theorem in section 7.8: if f is continuous on [a, b] and A(x) = ∫ from a to x of f(t) dt, then A′(x) = f(x). The JEE (Main) 2026 syllabus lists "the fundamental theorem of calculus" in Unit 8, and the JEE Advanced 2026 information brochure lists the "fundamental theorem of integral calculus" under Integral Calculus.
Neither syllabus prints the name Leibniz. The rule for variable limits is just that theorem joined to the chain rule, so it sits inside both. Check the syllabus for your own exam year.
What does the Leibniz rule say?
If f is continuous and u(x) and v(x) are differentiable, then
d/dx ∫ from u(x) to v(x) of f(t) dt = f(v(x)) · v′(x) − f(u(x)) · u′(x).
In words: put the upper limit into f and multiply by its derivative, then subtract the same thing for the lower limit. A constant limit has derivative zero, so its term disappears.
Why does it work?
Let F be an antiderivative of f, so F′ = f. By the second fundamental theorem, the integral equals F(v(x)) − F(u(x)). Differentiate with the chain rule: F′(v(x)) v′(x) − F′(u(x)) u′(x), which is f(v(x)) v′(x) − f(u(x)) u′(x).
You never had to know what F looks like, only that it exists. The first fundamental theorem guarantees that when f is continuous. That is why the rule works even for integrands like (sin t)/t that have no elementary antiderivative.
Worked example 1: one moving limit
Find d/dx ∫ from 0 to x² of cos t dt.
The upper limit is x², with derivative 2x. The lower limit is constant. So the answer is cos(x²) · 2x = 2x cos(x²).
Check: the integral is sin(x²), and its derivative is 2x cos(x²). Same answer.
Worked example 2: both limits move
Find d/dx ∫ from sin x to cos x of 1/(1 + t²) dt.
Upper term: 1/(1 + cos² x) · (−sin x).
Lower term: 1/(1 + sin² x) · cos x.
Answer: −sin x/(1 + cos² x) − cos x/(1 + sin² x).
Check: the integral is tan⁻¹(cos x) − tan⁻¹(sin x), and differentiating that gives the same two terms.
Worked example 3: a limit of the form 0/0
Find the limit as x → 0 of [∫ from 0 to x of sin(t²) dt] / x³.
Top and bottom both tend to 0. By L'Hospital's rule, which the JEE Advanced syllabus lists, differentiate each: the top gives sin(x²) by the Leibniz rule, the bottom gives 3x². Since sin(x²)/x² → 1, the limit is 1/3.
Worked example 4: x inside the integrand
Let g(x) = ∫ from 0 to x of (x − t) cos t dt. Find g′(x).
The tempting answer puts t = x into (x − t) cos t and gets 0. That is wrong, because x also appears inside the integrand. Pull it out first:
g(x) = x ∫ from 0 to x of cos t dt − ∫ from 0 to x of t cos t dt.
Now use the product rule on the first part and the Leibniz rule on both:
g′(x) = ∫ from 0 to x of cos t dt + x cos x − x cos x = sin x.
Check: working the integral out directly gives g(x) = 1 − cos x, whose derivative is sin x.
Worked example 5: maxima and minima
Let F(x) = ∫ from 0 to x of (t − 1)(t − 2) dt. Find its local maximum and minimum.
F′(x) = (x − 1)(x − 2), which is zero at x = 1 and x = 2. Then F″(x) = 2x − 3, so F″(1) = −1 (local maximum) and F″(2) = 1 (local minimum). If the question wants values, F(1) = 1/3 − 3/2 + 2 = 5/6 and F(2) = 8/3 − 6 + 4 = 2/3.
Which mistakes cost marks?
Dropping the chain factor. Writing cos(x²) instead of 2x cos(x²) in example 1.
Losing the minus sign on the lower-limit term, or forgetting that term when the lower limit also depends on x.
Leaving x inside the integrand. As example 4 shows, pull x outside the integral before you apply the rule.
Trying to integrate first. If the integrand has no simple antiderivative, the rule is the only route.
Ignoring continuity. The rule needs f to be continuous. With floor or modulus functions, split the interval first, as in our guide to definite integrals with modulus and floor.
If you keep losing the chain factor or the sign, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see where each term comes from.
A practice routine
Write the rule from memory at the start of each session.
For every problem, write the upper term and lower term on separate lines before simplifying.
Once per session, pick a problem you can integrate directly and check the rule against it, as in examples 1 and 2.
Mix in one limit question and one maxima question, since JEE often hides the rule inside them.
Keep an error log with three labels: chain factor, sign, x inside. Our post on definite integral properties pairs well with this set.
Tonight, take three past JEE questions with variable limits and write the upper and lower terms on separate lines before you simplify.
