# Leibniz Rule for Differentiating Integrals

> Leibniz rule for JEE Mathematics: differentiating integrals with variable limits, the nth-derivative product formula, plus mixed-scope problems solved.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/leibniz-rule-differentiation-integrals
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Leibniz Rule for Differentiating Integrals", PrepElephant, https://prepelephant.com/topics/jee/mathematics/leibniz-rule-differentiation-integrals

## Direct answer

Differentiating an integral whose limits move obeys the Leibniz rule: d/dx ∫ from u(x) to v(x) of f(t) dt = f(v)·v'(x) − f(u)·u'(x) — the integrand evaluated at each limit, times that limit's derivative, upper limit positive, lower limit negative. When x itself appears inside the integrand, an extra partial-derivative term integrates through, but JEE problems usually keep the integrand clean and let the limits carry all the x-dependence. The name Leibniz also attaches to the nth-derivative product rule, (uv)^(n) = Σ C(n, r)·u^(r)·v^(n−r), a completely separate result that shares the page in most Indian textbooks and appears in JEE under the same heading — both formulas are fair game and both reward the same discipline: track which variable is being differentiated and which is being integrated.

## What you must remember

- **Variable-limit formula:** d/dx ∫_{u(x)}^{v(x)} f(t)dt = f(v)·v' − f(u)·u'; constant limits give zero derivative only when the integrand is independent of x.
- **Sign convention:** upper limit contributes +, lower limit contributes −; reversing the limits flips the sign of everything.
- **Classic instance:** d/dx ∫₀^{x²} f(t)dt = 2x·f(x²); with f(t) = √(1 + t³), the derivative is 2x·√(1 + x⁶).
- **Both limits moving:** for ∫_{x}^{x²} f(t)dt the derivative is 2x·f(x²) − 1·f(x) — each limit handled independently.
- **Leibniz nth-derivative rule:** (uv)^(n) = Σ from r = 0 to n of C(n, r)·u^(r)·v^(n−r); for u = e^(ax) and v = sin bx, the sum telescopes into e^(ax) times a sinusoid with amplitude (a² + b²)^(n/2).
- **Newton–Leibniz connection:** the variable-limit rule is the first fundamental theorem of calculus composed with the chain rule — understanding this derivation replaces memorisation.
- **Inverse problems:** equations of the form ∫₀^{x} f(t)dt = f(x)·g(x) are solved by differentiating both sides with this rule and solving the resulting differential or functional condition.

## A moving upper limit

Let F(x) = ∫₀^{x²} √(1 + t³) dt. By the Leibniz rule with v(x) = x², v'(x) = 2x: F'(x) = √(1 + (x²)³)·2x = 2x√(1 + x⁶). No antiderivative of √(1 + t³) is needed — that is the entire point, and the reason this rule appears in every serious paper: it extracts derivative information from integrals that cannot be evaluated in closed form. Extend the same function: F''(x) requires the product rule on 2x√(1 + x⁶), giving 2√(1 + x⁶) + 2x·(1/2)(1 + x⁶)^(−1/2)·6x⁵ = 2√(1 + x⁶) + 6x⁶/√(1 + x⁶).

The nth-derivative half of the page runs on the product formula. Compute the 4th derivative of x²e^(2x)? With u = e^(2x), v = x²: (uv)'''' = e^(2x)[2⁴x² + 4·2³·2x + 6·2²·2 + 0] = e^(2x)(16x² + 64x + 48), reading C(4,0)x²·2⁴ + C(4,1)(2x)·2³ + C(4,2)(2)·2²; higher derivatives of x² vanish after the second, truncating the sum. The binomial-coefficient structure is why the formula is called Leibniz: differentiation of a product behaves like raising to a power.

## Variable-limit traps

The dominant error is evaluating the integrand at x instead of at the limit: for ∫₀^{x²} f(t)dt, the derivative contains f(x²), not f(x) — the chain rule enters through the limit, and every wrong option in a multiple-choice set is built on the unchained version. Second, dropping the sign for a decreasing lower limit: if the lower limit is 1/x, its derivative −1/x² enters as −f(1/x)·(−1/x²) = +f(1/x)/x². Third, in the nth-derivative rule, miscounting which function's derivative is taken — the r-th derivative lands on u and the (n − r)-th on v, exactly as in the binomial expansion. Main-level questions compute a value at a point (evaluate F'(1) for a given f); Advanced build equations — find f given ∫₀^x f(t)dt = x·f(x) type functional-integral conditions, where differentiating via Leibniz is the opening move and the resulting differential equation is the rest of the solution.

## Frequently asked questions

### What is the Leibniz rule for differentiating an integral with variable limits?

d/dx ∫_{u(x)}^{v(x)} f(t)dt = f(v)·v'(x) − f(u)·u'(x): evaluate the integrand at each moving limit and weight by that limit's derivative, upper positive, lower negative.

### Why is the integrand evaluated at x² and not x for ∫₀^{x²} f(t)dt?

Because the chain rule acts through the limit: the upper limit v = x² has v' = 2x, so the derivative is f(x²)·2x — the substitution of the limit into the integrand is mandatory.

### What is the Leibniz rule for nth derivatives of a product?

(uv)^(n) = Σ C(n, r)·u^(r)·v^(n−r), a binomial-pattern sum; it truncates early whenever one factor is a polynomial of low degree.

### What happens when both limits depend on x?

Each limit contributes independently: d/dx ∫_{x}^{x²} f(t)dt = 2x·f(x²) − f(x), keeping the signs straight by the upper-positive, lower-negative convention.

### Can this rule differentiate an integral that cannot be evaluated?

Yes — that is its chief power: F(x) = ∫₀^{x²} √(1 + t³)dt has derivative 2x√(1 + x⁶) even though no elementary antiderivative of the integrand exists.