Leibniz Rule for Differentiating Integrals
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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.