How do I combine Leibnitz's rule with L'Hopital's rule?

When a limit involves an integral with a variable upper limit in a 0/0 or infinity/infinity form, apply L'Hopital and differentiate the integral using Leibnitz. This is standard for JEE limit-of-integral problems.

Can I use Leibnitz's rule when the upper limit is infinity?

Yes, provided the integral converges and appropriate convergence conditions hold. For JEE problems with infinity as a limit, the rule usually applies straightforwardly after checking convergence.

Differentiation Under the Integral Sign

Q: What is Leibnitz's rule in one sentence?

Leibnitz's rule says the derivative of the integral of f(t) from g(x) to h(x) with respect to x equals f(h(x)) times h prime of x minus f(g(x)) times g prime of x.

Q: When do I need Leibnitz's rule instead of Newton-Leibnitz?

Newton-Leibnitz evaluates a definite integral to a number. Leibnitz's rule differentiates a definite integral whose limits are functions of x. Use it whenever you want the derivative of an integral or when the integrand has no closed-form antiderivative.

Q: Does Leibnitz's rule work if f has discontinuities?

The standard rule requires f to be continuous. If f has a jump inside, split the integral at the discontinuity first, then apply Leibnitz to each piece.

Q: What if both limits and the integrand depend on the variable?

Use the full form: differentiate the limits AND add the integral of the partial derivative of the integrand. Both effects contribute to the total derivative.

Q: How does Leibnitz's rule prove the second Fundamental Theorem of Calculus?

The second FTC is the special case with g(x) equal to a constant and h(x) equal to x: the rule reduces to f(x) times 1 minus f(a) times 0, which equals f(x).

Q: What's a good sanity check when applying Leibnitz's rule?

Verify boundary behaviour. At x equal to the lower limit, the integral is zero, and its derivative should match f evaluated there. Also check that a constant upper limit yields derivative zero.

MathsIntegralsFor JEE aspirants

Differentiation Under the Integral Sign (DUIS), also called Leibnitz's Rule, lets you differentiate a definite integral whose limits are functions of $x$ : if $F (x) = \int_{g (x)}^{h (x)} f (t) d t$ , then $F^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$ . This rule is a JEE Advanced favourite because it turns "find the derivative of this integral" problems into direct one-line substitutions, and it also handles integrals with a parameter inside the integrand (the "modified" Leibnitz rule). Mastering both variants covers every DUIS question likely to appear in the exam.

Key Formulas - Quick Reference

Leibnitz's rule (variable limits): $\frac{d}{d x} \int_{g (x)}^{h (x)} f (t) d t = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$
Simple form (upper limit variable): $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$
Modified Leibnitz rule (parameter in integrand): $\frac{d}{d α} \int_{a}^{b} f (x, α) d x = \int_{a}^{b} \frac{\partial f}{\partial α} (x, α) d x$
Full Leibnitz (both limits and integrand depend on $α$ ): $\frac{d}{d α} \int_{g (α)}^{h (α)} f (x, α) d x = f (h (α), α) h^{'} (α) - f (g (α), α) g^{'} (α) + \int_{g (α)}^{h (α)} \frac{\partial f}{\partial α} d x$
L'Hôpital combined with Leibnitz: use Leibnitz to differentiate integrals with variable upper limits in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ limits

1. Leibnitz's Rule (Variable Limits)

Statement: If

f

is continuous on

[a, b]

and

g (x)

h (x)

are differentiable functions whose values lie in

[a, b]

, then

\frac{d}{d x} \int_{g (x)}^{h (x)} f (t) d t = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)

Why the rule works (intuition and proof)

Let $P (t)$ be an antiderivative of $f (t)$ , so $P^{'} (t) = f (t)$ . By the Fundamental Theorem:

F (x) = \int_{g (x)}^{h (x)} f (t) d t = P (h (x)) - P (g (x))

Differentiate both sides using the chain rule:

F^{'} (x) = P^{'} (h (x)) h^{'} (x) - P^{'} (g (x)) g^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)

Notice that $F^{'} (x)$ depends only on the endpoint values of $f$ and the derivatives of the limits, never on the antiderivative itself. This is what makes the rule so powerful.

Simple Cases

Upper limit is $x$ , lower is a constant: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$
Lower limit is $x$ , upper is a constant: $\frac{d}{d x} \int_{x}^{b} f (t) d t = - f (x)$
Upper limit is $u (x)$ , lower is a constant: $\frac{d}{d x} \int_{a}^{u (x)} f (t) d t = f (u (x)) u^{'} (x)$

Figure 1: The integral

F (x) = \int_{g (x)}^{h (x)} f (t) d t

is the shaded area whose derivative comes from the endpoints, not the interior.

2. Solved Examples: Variable Limits

Solved Example 1

F (x) = \int_{x}^{x^{2}} sin t d t

, find

F^{'} (x)

Solution:

By Leibnitz's rule with $g (x) = x$ , $h (x) = x^{2}$ , $f (t) = sin t$ :

F^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x) = sin (x^{2}) \cdot 2 x - sin x \cdot 1

F^{'} (x) = 2 x sin x^{2} - sin x

Solved Example 2

f (x) = cos x - \int_{0}^{x} (x - t) f (t) d t

, show that

f^{''} (x) + f (x) = - cos x

Solution:

Rewrite: $f (x) = cos x - x \int_{0}^{x} f (t) d t + \int_{0}^{x} t f (t) d t$ .

Differentiate using the product rule and Leibnitz:

f^{'} (x) = - sin x - (\int_{0}^{x} f (t) d t + x \cdot f (x)) + x \cdot f (x)

= - sin x - \int_{0}^{x} f (t) d t

Differentiate again:

f^{''} (x) = - cos x - f (x)

Therefore $f^{''} (x) + f (x) = - cos x$ , as required.

Solved Example 3

F (x) = \int_{e^{2 x}}^{e^{3 x}} \frac{t}{lo g _{e} t} d t

, find the first and second derivatives of

F (x)

with respect to

ln x

x = ln 2

Solution:

Apply Leibnitz's rule:

\frac{d F}{d x} = \frac{e ^{3 x}}{ln e ^{3 x}} \cdot 3 e^{3 x} - \frac{e ^{2 x}}{ln e ^{2 x}} \cdot 2 e^{2 x} = \frac{e ^{3 x}}{3 x} \cdot 3 e^{3 x} - \frac{e ^{2 x}}{2 x} \cdot 2 e^{2 x} = \frac{e ^{6 x} - e ^{4 x}}{x}

Now $\frac{d F}{d ( ln x )} = \frac{d F}{d x} \cdot \frac{d x}{d ( ln x )} = \frac{d F}{d x} \cdot x = e^{6 x} - e^{4 x}$ .

Second derivative:

\frac{d ^{2} F}{d ( ln x ) ^{2}} = \frac{d}{d ( ln x )} (e^{6 x} - e^{4 x}) = \frac{d}{d x} (e^{6 x} - e^{4 x}) \cdot x = (6 e^{6 x} - 4 e^{4 x}) x

Evaluate at $x = ln 2$ (so $e^{x} = 2$ ):

First derivative $= e^{6 x} - e^{4 x} = 2^{6} - 2^{4} = 64 - 16 = 48$
Second derivative $= (6 \cdot 2^{6} - 4 \cdot 2^{4}) \cdot ln 2 = (384 - 64) ln 2 = 320 ln 2 = 5 \cdot 2^{6} ln 2$

Solved Example 4

Evaluate

x \to \infty lim \frac{( \int _{0}^{x} e ^{t^{2}} d t ) ^{2}}{\int _{0}^{x} e ^{2 t^{2}} d t}

Solution:

This is a $\frac{\infty}{\infty}$ form. Apply L'Hôpital using Leibnitz to differentiate the integrals:

x \to \infty lim \frac{2 \int _{0}^{x} e ^{t^{2}} d t \cdot e ^{x^{2}}}{e ^{2 x^{2}}} = x \to \infty lim \frac{2 \int _{0}^{x} e ^{t^{2}} d t}{e ^{x^{2}}}

Still $\infty/\infty$ . Apply L'Hôpital again:

= x \to \infty lim \frac{2 e ^{x^{2}}}{2 x \cdot e ^{x^{2}}} = x \to \infty lim \frac{1}{x} = 0

Solved Example 5

ϕ (x) = cos x - \int_{0}^{x} (x - t) ϕ (t) d t

, find

ϕ^{''} (x) + ϕ (x)

Solution:

This is the same setup as Solved Example 2 with $ϕ$ in place of $f$ . The identical calculation gives:

ϕ^{''} (x) + ϕ (x) = - cos x

Solved Example 6

Determine a positive integer

n \leq 5

such that

\int_{0}^{1} e^{x} (x - 1)^{n} d x = 16 - 6 e

Solution:

Let $I_{n} = \int_{0}^{1} e^{x} (x - 1)^{n} d x$ . Integrate by parts with $u = (x - 1)^{n}$ , $d v = e^{x} d x$ :

I_{n} = [e^{x} (x - 1)^{n}]_{0}^{1} - n \int_{0}^{1} e^{x} (x - 1)^{n - 1} d x

= (0) - (- 1)^{n} - n I_{n - 1} = (- 1)^{n + 1} - n I_{n - 1}

Compute base case:

I_{1} = \int_{0}^{1} e^{x} (x - 1) d x = [(x - 1) e^{x}]_{0}^{1} - \int_{0}^{1} e^{x} d x = 0 - (- 1) - (e - 1) = 2 - e

Recurrence:

$I_{2} = (- 1)^{3} - 2 I_{1} = - 1 - 2 (2 - e) = - 5 + 2 e$
$I_{3} = (- 1)^{4} - 3 I_{2} = 1 - 3 (- 5 + 2 e) = 1 + 15 - 6 e = 16 - 6 e$ ✓

Hence $n = 3$ .

3. Modified Leibnitz Rule (Parameter in Integrand)

If the integrand depends on a parameter

α

but the limits do not, then

\frac{d}{d α} \int_{a}^{b} f (x, α) d x = \int_{a}^{b} \frac{\partial f}{\partial α} (x, α) d x

This is the "modified Leibnitz rule" or Feynman's trick: differentiate under the integral sign to introduce a new variable, solve the resulting ODE for the parametrised integral, and back-substitute.

Solved Example 7

Evaluate

I (b) = \int_{0}^{1} \frac{x ^{b} - 1}{ln x} d x

, where

b

is a parameter.

Solution:

Differentiate with respect to $b$ using the modified Leibnitz rule:

\frac{d I}{d b} = \int_{0}^{1} \frac{\partial}{\partial b} (\frac{x ^{b} - 1}{ln x}) d x = \int_{0}^{1} \frac{x ^{b} ln x}{ln x} d x = \int_{0}^{1} x^{b} d x = \frac{1}{b + 1}

Integrate w.r.t. $b$ : $I (b) = ln (b + 1) + C$ .

At $b = 0$ : $I (0) = \int_{0}^{1} \frac{x ^{0} - 1}{ln x} d x = 0$ , so $C = 0$ .

Therefore $I (b) = ln (b + 1)$ .

Solved Example 8

Evaluate

I (a) = \int_{0}^{1} \frac{tan ^{- 1} ( a x )}{x 1 - x ^{2}} d x

, where

a

is a parameter.

Solution:

Differentiate with respect to $a$ :

\frac{d I}{d a} = \int_{0}^{1} \frac{x}{( 1 + a ^{2} x ^{2} )} \cdot \frac{1}{x 1 - x ^{2}} d x = \int_{0}^{1} \frac{d x}{( 1 + a ^{2} x ^{2} ) 1 - x ^{2}}

Substitute $x = sin t$ , $d x = cos t d t$ . Limits: $x = 0 \to t = 0$ ; $x = 1 \to t = π /2$ .

\frac{d I}{d a} = \int_{0}^{π /2} \frac{cos t d t}{( 1 + a ^{2} sin ^{2} t ) cos t} = \int_{0}^{π /2} \frac{d t}{1 + a ^{2} sin ^{2} t}

Divide numerator and denominator by $cos^{2} t$ :

= \int_{0}^{π /2} \frac{sec ^{2} t d t}{sec ^{2} t + a ^{2} tan ^{2} t} = \int_{0}^{π /2} \frac{sec ^{2} t d t}{1 + ( 1 + a ^{2} ) tan ^{2} t}

Let $u = tan t$ . This gives $\frac{d I}{d a} = \frac{1}{1 + a ^{2}} \cdot \frac{π}{2}$ .

Integrate: $I (a) = \frac{π}{2} ln (a + 1 + a^{2}) + C$ .

At $a = 0$ : $I (0) = 0$ , so $C = 0$ . Hence $I (a) = \frac{π}{2} ln (a + 1 + a^{2})$ .

Solved Example 9

Evaluate

x \to 0 lim \frac{\int _{0}^{x^{2}} cos t ^{2} d t}{x sin x}

Solution:

This is a $\frac{0}{0}$ form. Apply L'Hôpital, using Leibnitz on the numerator:

Numerator derivative: $\frac{d}{d x} \int_{0}^{x^{2}} cos t^{2} d t = cos (x^{4}) \cdot 2 x$

Denominator derivative: $\frac{d}{d x} (x sin x) = sin x + x cos x$

x \to 0 lim \frac{2 x cos ( x ^{4} )}{sin x + x cos x} = x \to 0 lim \frac{2 cos ( x ^{4} )}{\frac{s i n x}{x} + cos x} = \frac{2 \cdot 1}{1 + 1} = 1

Solved Example 10

f (x) = \int_{l o g_{e} x}^{x} \frac{d t}{x + t}

, find

f^{'} (x)

Solution:

Both the limits AND the integrand depend on $x$ . Use the full Leibnitz rule (both effects sum):

f^{'} (x) = limit effect \frac{1}{x + x} \cdot 1 - \frac{1}{x + ln x} \cdot \frac{1}{x} + parameter effect \int_{l n x}^{x} \frac{\partial}{\partial x} (\frac{1}{x + t}) d t

The parameter integral: $\frac{\partial}{\partial x} \frac{1}{x + t} = - \frac{1}{( x + t ) ^{2}}$ , so

\int_{l n x}^{x} - \frac{d t}{( x + t ) ^{2}} = [\frac{1}{x + t}]_{l n x}^{x} = \frac{1}{2 x} - \frac{1}{x + ln x}

Total:

f^{'} (x) = \frac{1}{2 x} - \frac{1}{x ( x + ln x )} + \frac{1}{2 x} - \frac{1}{x + ln x} = \frac{1}{x} - \frac{x + 1}{x ( x + ln x )} = \frac{ln x - 1}{x ( x + ln x )}

Common Mistakes to Avoid

Watch out

Forgetting to multiply by $h^{'} (x)$ or $g^{'} (x)$ . Writing $\frac{d}{d x} \int_{0}^{x^{2}} f (t) d t = f (x^{2})$ is wrong; the correct answer is $f (x^{2}) \cdot 2 x$ .
Dropping the minus sign for the lower limit. If the lower limit is $g (x)$ , its contribution is $- f (g (x)) g^{'} (x)$ , not $+ f (g (x)) g^{'} (x)$ .
Using Leibnitz on an integral where the integrand also depends on $x$ but treating the integrand as a constant. If $F (x) = \int_{a}^{h (x)} f (x, t) d t$ , you must add the parameter term $\int_{a}^{h (x)} \frac{\partial f}{\partial x} d t$ .
Applying the rule when the integrand is discontinuous inside the interval. Leibnitz requires continuity of $f$ on the range spanned by the limits.
In the modified rule, forgetting to fix the constant of integration using a known value of the parametrised integral (like $I (0) = 0$ ).
Confusing $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ with $\int_{a}^{x} f^{'} (t) d t = f (x) - f (a)$ . These are the two halves of the Fundamental Theorem, not the same statement.

Frequently Asked Questions

Q1. What is Leibnitz's rule in one sentence?

Leibnitz's rule says the derivative of $\int_{g (x)}^{h (x)} f (t) d t$ with respect to $x$ equals $f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$ : the integrand evaluated at each limit, times the derivative of that limit, with the lower one carrying a minus sign.

Q2. When do I need Leibnitz's rule instead of Newton-Leibnitz?

Newton-Leibnitz evaluates a definite integral to a number using an antiderivative. Leibnitz's rule differentiates a definite integral (whose limits are functions of $x$ ) with respect to $x$ . Use it whenever you're asked for $F^{'} (x)$ where $F$ is defined as an integral, or when the integrand has no closed-form antiderivative but you can still find $F^{'} (x)$ by the rule.

Q3. What is the modified Leibnitz rule and when is it useful?

If the limits are fixed but the integrand depends on a parameter $α$ , then $\frac{d}{d α} \int_{a}^{b} f (x, α) d x = \int_{a}^{b} \frac{\partial f}{\partial α} d x$ . This "Feynman trick" is used to evaluate integrals that resist standard techniques: differentiate with respect to a parameter, evaluate the simpler derivative, then integrate the result back.

Q4. How do I combine Leibnitz's rule with L'Hôpital's rule?

When a limit involves an integral with a variable upper limit in a $0/0$ or $\infty/\infty$ form, use L'Hôpital and differentiate the integral using Leibnitz. This is a standard JEE trick for limits like $lim_{x \to 0} \frac{\int _{0}^{x} f ( t ) d t}{x}$ , which equals $f (0)$ by one application.

Q5. Does Leibnitz's rule work if $f$ has discontinuities?

The standard statement requires $f$ to be continuous on the range spanned by the limits. If $f$ has a jump inside, split the integral at the discontinuity first, then apply Leibnitz to each piece. For an unbounded discontinuity, the integral becomes improper and needs its own analysis.

Q6. What if both limits and the integrand depend on the variable?

Use the full form: differentiate the limits (Leibnitz variable-limit part) AND add the integral of the partial derivative of the integrand (modified Leibnitz part). The total is $f (h, x) h^{'} - f (g, x) g^{'} + \int_{g}^{h} \frac{\partial f}{\partial x} d t$ .

Q7. Can I use Leibnitz's rule when the upper limit is $\infty$ ?

Yes, provided the integral converges and appropriate convergence conditions hold (uniform convergence for the modified rule). For JEE problems with $\infty$ as a limit, verify convergence first; usually the rule applies straightforwardly.

Q8. How does Leibnitz's rule prove the second Fundamental Theorem of Calculus?

The second FTC states $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ , which is the special case of Leibnitz's rule with $g (x) = a$ (constant, so $g^{'} (x) = 0$ ) and $h (x) = x$ (so $h^{'} (x) = 1$ ): the rule reduces to $f (x) \cdot 1 - f (a) \cdot 0 = f (x)$ .

Q9. What's a good sanity check when applying Leibnitz's rule?

Verify units and boundary behaviour. If the integral is $\int_{0}^{x} f (t) d t$ , then at $x = 0$ the integral is zero, and its derivative $f (x)$ should equal $f (0)$ at $x = 0$ . Also check that a constant upper limit yields derivative zero (as it should).

Previous year questions on Differentiation Under the Integral Sign

4 questions from past papers, each with a step-by-step solution.

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1. Leibnitz's Rule (Variable Limits)

Why the rule works (intuition and proof)

Simple Cases

2. Solved Examples: Variable Limits

3. Modified Leibnitz Rule (Parameter in Integrand)

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is Leibnitz's rule in one sentence?

Q2. When do I need Leibnitz's rule instead of Newton-Leibnitz?

Q3. What is the modified Leibnitz rule and when is it useful?

Q4. How do I combine Leibnitz's rule with L'Hôpital's rule?

Q5. Does Leibnitz's rule work if f has discontinuities?

Q6. What if both limits and the integrand depend on the variable?

Q7. Can I use Leibnitz's rule when the upper limit is ∞?

Q8. How does Leibnitz's rule prove the second Fundamental Theorem of Calculus?

Q9. What's a good sanity check when applying Leibnitz's rule?

Previous year questions on Differentiation Under the Integral Sign

Ready to master Integrals?

Q5. Does Leibnitz's rule work if $f$ has discontinuities?

Q7. Can I use Leibnitz's rule when the upper limit is $\infty$ ?