General Leibniz Rule: Formula, Proof and Examples

What Is The General Leibniz Rule?

The general Leibniz rule is the formula for the nth derivative of a product of two functions: it adds up every way of sharing $n$ differentiations between the two factors, weighting each share by a binomial coefficient. It is the product rule extended from one derivative to $n$, and many textbooks call it Leibnitz theorem or the Leibniz formula.

Statement. Let $f$ and $g$ be functions that are $n$ times differentiable at $x$, where $n$ is a positive whole number. Then $fg$ is $n$ times differentiable at $x$, and

$$(fg)^{(n)}(x) = \sum_{k=0}^{n}\binom{n}{k},f^{(n-k)}(x),g^{(k)}(x)$$

The binomial coefficient counts the ways to choose $k$ items from $n$. It can be computed without factorials as

$$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{1\cdot 2\cdot 3\cdots k}, \qquad \binom{n}{0} = 1$$

so $\binom{4}{2} = \frac{4\cdot 3}{1\cdot 2} = 6$, and $\binom{n}{2} = \frac{n(n-1)}{2}$. Because the sum is symmetric, you may also write it as $\sum_{k=0}^{n}\binom{n}{k}f^{(k)}g^{(n-k)}$: same terms, listed in the opposite order.

Why Does The Leibniz Rule Look Like The Binomial Theorem?

The general Leibniz rule looks like the binomial theorem because both count the same thing: the ways to share $n$ identical steps between two places. Expanding $(a + b)^n$ gives $\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$; the Leibniz rule has the same coefficients, with powers replaced by orders of derivatives.

Write out the first few orders and the pattern is plain:

$$(fg)' = f'g + fg'$$

$$(fg)'' = f''g + 2f'g' + fg''$$

$$(fg)''' = f'''g + 3f''g' + 3f'g'' + fg'''$$

$$(fg)^{(4)} = f^{(4)}g + 4f'''g' + 6f''g'' + 4f'g''' + fg^{(4)}$$

The coefficients $1, 2, 1$ then $1, 3, 3, 1$ then $1, 4, 6, 4, 1$ are rows of Pascal's triangle. The reason is the same as in algebra. Each time you differentiate, every term splits into two: one where the new derivative lands on $f$, one where it lands on $g$. After $n$ rounds, the number of paths that put exactly $k$ derivatives on $g$ is $\binom{n}{k}$.

How Do You Prove The General Leibniz Rule?

The standard proof uses mathematical induction on $n$, and its key step is Pascal's rule for binomial coefficients. A Quora thread asks for exactly this, the proof of the Leibnitz rule for finding an nth derivative, so here is the sketch in three moves.

1. Base case, $n = 1$. The formula reads $(fg)' = \binom{1}{0}f'g + \binom{1}{1}fg' = f'g + fg'$, which is the product rule.

2. Inductive hypothesis. Assume the formula holds for some $n$: $(fg)^{(n)} = \sum_{k=0}^{n}\binom{n}{k}f^{(n-k)}g^{(k)}$.

3. Inductive step. Differentiate both sides once more, applying the product rule to every term:

$$(fg)^{(n+1)} = \sum_{k=0}^{n}\binom{n}{k}\left[f^{(n+1-k)}g^{(k)} + f^{(n-k)}g^{(k+1)}\right]$$

In the second group, rename $k + 1$ as $k$, then collect the terms that carry the same pair of derivative orders. The term $f^{(n+1-k)}g^{(k)}$ now has coefficient $\binom{n}{k} + \binom{n}{k-1}$, and Pascal's rule says

$$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$$

The two end terms, $k = 0$ and $k = n + 1$, have coefficient $1$, which matches $\binom{n+1}{0}$ and $\binom{n+1}{n+1}$. So

$$(fg)^{(n+1)} = \sum_{k=0}^{n+1}\binom{n+1}{k}f^{(n+1-k)}g^{(k)}$$

which is the formula for $n + 1$. By induction, it holds for every positive whole number $n$.

Can You Prove It Without Induction?

Another Quora thread asks how to prove the Leibniz theorem without mathematical induction, and the counting argument from the previous section is one way. Think of $\frac{d}{dx}$ acting on $fg$ as the sum of two operators, one that differentiates only $f$ and one that differentiates only $g$. These two commute, so the binomial theorem applies to their sum raised to the power $n$, and the coefficients come out as $\binom{n}{k}$ directly. Making "the operators commute" fully rigorous takes some care, which is why most courses teach the induction proof.

How Do You Use The General Leibniz Rule Step By Step?

You use the general Leibniz rule by choosing which factor to differentiate many times, listing the derivatives of each factor, and keeping only the terms that do not vanish. In practice the method is four steps.

1. Choose the factors. Let $f$ be the factor whose derivatives eventually become $0$ (a polynomial), and $g$ the factor with a clean nth-derivative pattern ($e^{ax}$, $\sin x$, $\ln x$).

2. List the derivatives. Write $f$, $f'$, $f''$, and so on until they reach $0$; write the general formula for $g^{(m)}$.

3. Write only the surviving terms. If $f$ is a polynomial of degree $d$, only $d + 1$ terms survive, however large $n$ is.

4. Insert the binomial coefficients and simplify. Pull out common factors last.

For step 2 you will reuse a few standard patterns, from the derivatives of exponential functions and their logarithmic and trigonometric cousins:

The sine and cosine rows hold with $x$ in radians. The $\ln x$ row gives $\frac{1}{x}$, $-\frac{1}{x^2}$, $\frac{2}{x^3}$, $-\frac{6}{x^4}$ for $m = 1$ to $4$, and the derivatives of logarithmic functions page shows where the first one comes from.

Examples Of The General Leibniz Rule

These six examples move from a single order to a recurrence used in series work.

Example 1

Find the third derivative of $x^2e^x$, then the nth derivative.

Take $f = x^2$, so $f' = 2x$, $f'' = 2$ and $f''' = 0$; take $g = e^x$, so every $g^{(m)} = e^x$. With $f$ placed as the factor that receives $k$ derivatives:

$$\left(x^2e^x\right)''' = \binom{3}{0}x^2e^x + \binom{3}{1}(2x)e^x + \binom{3}{2}(2)e^x$$

$$= \left(x^2 + 6x + 6\right)e^x$$

For general $n$, the same three terms survive:

$$\left(x^2e^x\right)^{(n)} = x^2e^x + n(2x)e^x + \frac{n(n-1)}{2}(2)e^x$$

$$= \left(x^2 + 2nx + n(n-1)\right)e^x$$

With $n = 20$ this gives $\left(x^2 + 40x + 380\right)e^x$, the result quoted at the top.

Final answer: $\left(x^2e^x\right)''' = \left(x^2 + 6x + 6\right)e^x$ and $\left(x^2e^x\right)^{(n)} = \left(x^2 + 2nx + n(n-1)\right)e^x$.

Example 2

Find the fourth derivative of $x\sin x$.

The tempting version of the rule copies the Pascal-free pattern, adding each split of the derivatives once without coefficients:

$$\sum_{k=0}^{4}f^{(k)}g^{(4-k)} \quad\text{with } f = x,\ g = \sin x$$

$$= x\sin x + 1\cdot(-\cos x) = x\sin x - \cos x$$

Test that version on a product with a known answer. For $x \cdot x = x^2$ and $n = 2$, it gives $x\cdot 0 + 1\cdot 1 + 0\cdot x = 1$, but $\left(x^2\right)'' = 2$. The version without coefficients fails, because the middle term $f'g'$ appears in two ways and must be counted twice.

With the coefficients, only $k = 0$ and $k = 1$ survive since $f'' = 0$:

$$(x\sin x)^{(4)} = \binom{4}{0}x,(\sin x)^{(4)} + \binom{4}{1}(1),(\sin x)'''$$

$$= x\sin x + 4(-\cos x)$$

$$= x\sin x - 4\cos x$$

At $x = 0$ the correct value is $-4$, while the coefficient-free version gives $-1$.

Final answer: $(x\sin x)^{(4)} = x\sin x - 4\cos x$.

Example 3

Find the fourth derivative of $x^3\ln x$, for $x > 0$.

Let the polynomial $x^3$ take $k$ derivatives: $3x^2$, $6x$, $6$, then $0$. The logarithm takes the rest, using $(\ln x)' = \frac{1}{x}$, $(\ln x)'' = -\frac{1}{x^2}$, $(\ln x)''' = \frac{2}{x^3}$, $(\ln x)^{(4)} = -\frac{6}{x^4}$.

$$k = 0: \quad \binom{4}{0}x^3\left(-\frac{6}{x^4}\right) = -\frac{6}{x}$$

$$k = 1: \quad \binom{4}{1}\left(3x^2\right)\left(\frac{2}{x^3}\right) = \frac{24}{x}$$

$$k = 2: \quad \binom{4}{2}(6x)\left(-\frac{1}{x^2}\right) = -\frac{36}{x}$$

$$k = 3: \quad \binom{4}{3}(6)\left(\frac{1}{x}\right) = \frac{24}{x}$$

$$\left(x^3\ln x\right)^{(4)} = \frac{-6 + 24 - 36 + 24}{x} = \frac{6}{x}$$

The usual first instinct is to give $x^3$ all four derivatives and stop; the result $0\cdot\ln x$ looks tidy but drops every term that actually survives.

Final answer: $\left(x^3\ln x\right)^{(4)} = \frac{6}{x}$.

Example 4

Find the nth derivative of $x^2e^{2x}$.

This is the function in a Quora question on using Leibniz's rule for the nth derivative. Here $\left(e^{2x}\right)^{(m)} = 2^m e^{2x}$, and $x^2$ gives three surviving terms:

$$\left(x^2e^{2x}\right)^{(n)} = x^2\cdot 2^n e^{2x} + n(2x),2^{n-1}e^{2x} + \frac{n(n-1)}{2}(2),2^{n-2}e^{2x}$$

Factor out $2^{n-2}e^{2x}$:

$$= 2^{n-2}e^{2x}\left(4x^2 + 4nx + n(n-1)\right)$$

Check with $n = 1$: $2^{-1}e^{2x}\left(4x^2 + 4x\right) = \left(2x^2 + 2x\right)e^{2x}$, which matches the product rule.

Final answer: $\left(x^2e^{2x}\right)^{(n)} = 2^{n-2}e^{2x}\left(4x^2 + 4nx + n(n-1)\right)$.

Example 5

Find the fifth derivative of $x^2\ln x$, for $x > 0$.

A Brainly question asks for the general nth derivative of $y = x^2\log x$, and the fifth is a good way to see its shape. The derivatives of $\ln x$ needed are $(\ln x)^{(5)} = \frac{24}{x^5}$, $(\ln x)^{(4)} = -\frac{6}{x^4}$ and $(\ln x)''' = \frac{2}{x^3}$.

$$k = 0: \quad x^2\cdot\frac{24}{x^5} = \frac{24}{x^3}$$

$$k = 1: \quad \binom{5}{1}(2x)\left(-\frac{6}{x^4}\right) = -\frac{60}{x^3}$$

$$k = 2: \quad \binom{5}{2}(2)\left(\frac{2}{x^3}\right) = \frac{40}{x^3}$$

$$\left(x^2\ln x\right)^{(5)} = \frac{24 - 60 + 40}{x^3} = \frac{4}{x^3}$$

Final answer: $\left(x^2\ln x\right)^{(5)} = \frac{4}{x^3}$.

Example 6

For $y = \sin^{-1}x$, show that $\left(1 - x^2\right)y_{n+2} - (2n+1)x,y_{n+1} - n^2y_n = 0$, where $y_n$ is the nth derivative, and use it to find $y^{(5)}(0)$.

First derivatives, for $-1 < x < 1$, from the derivatives of inverse trigonometric functions:

$$y_1 = \frac{1}{\sqrt{1 - x^2}}$$

$$\left(1 - x^2\right)y_1^2 = 1$$

Differentiate, then divide by $2y_1$:

$$\left(1 - x^2\right)y_2 - x,y_1 = 0$$

Apply the general Leibniz rule $n$ times. In $\left[\left(1 - x^2\right)y_2\right]^{(n)}$, the factor $1 - x^2$ has derivatives $-2x$, $-2$, then $0$, so three terms survive; in $\left[x,y_1\right]^{(n)}$, two survive:

$$\left(1 - x^2\right)y_{n+2} - 2nx,y_{n+1} - n(n-1)y_n - x,y_{n+1} - n,y_n = 0$$

$$\left(1 - x^2\right)y_{n+2} - (2n+1)x,y_{n+1} - n^2y_n = 0$$

At $x = 0$ the recurrence reduces to $y_{n+2}(0) = n^2y_n(0)$. Starting from $y_1(0) = 1$:

$$y_3(0) = 1^2\cdot y_1(0) = 1$$

$$y_5(0) = 3^2\cdot y_3(0) = 9$$

Final answer: the recurrence holds for every $n$, and $y^{(5)}(0) = 9$.

Where Is The General Leibniz Rule Used?

The general Leibniz rule is used whenever a high-order derivative of a product is needed and repeated product-rule work would be impractical:

• Series coefficients. The Taylor and Maclaurin series need $f^{(n)}(0)$ for every $n$. Recurrences like the one in Example 6 produce all of them at once, which is how the series for $\sin^{-1}x$ is usually built.

• Differential equations. Differentiating an equation such as $\left(1 - x^2\right)y'' - xy' + a^2y = 0$ $n$ times gives a recurrence for the coefficients of a power-series solution, and the Leibniz rule supplies the coefficients.

• Special functions. Rodrigues' formula writes each Legendre polynomial using the nth derivative of $\left(x^2 - 1\right)^n = (x-1)^n(x+1)^n$, a product the Leibniz rule expands term by term.

• Exam problems. Questions of the form "find $y_n$" or "prove that $\left(1 - x^2\right)y_{n+2} - \ldots = 0$" are a standard final step in differential calculus courses.

Why Did Calculus Need A Rule For The Nth Derivative?

The rule exists because "differentiating a product again and again doubles the work every time". One product rule turns one term into two; doing it twice gives four, then eight. Without a way to collect the matching terms, the 10th derivative of $x^2e^x$ would mean tracking 1,024 pieces of algebra.

• The bookkeeping is the problem. Most of those 1,024 pieces are equal or zero. The Leibniz rule groups the equal ones (that is what $\binom{n}{k}$ counts) and drops the zero ones before they are ever written.

• Notation made it visible. Gottfried Leibniz's 1684 paper in Acta Eruditorum set out the product rule with his $d$ notation. Wikipedia's product rule article notes that the discovery is credited to Leibniz, while one translator of his papers, J. M. Child, argued that it is due to Isaac Barrow.

• Series needed every derivative. Once mathematicians wanted to write functions as power series, they needed every derivative at a point, not just the first two.

What Are The Most Common Mistakes With The General Leibniz Rule?

Most errors with the Leibniz rule happen before any differentiation: the wrong factor is chosen, a coefficient is dropped, or a sign is lost in the logarithm.

Mistake 1: Giving The Many Derivatives To The Wrong Factor

Where it slips in:

Products of a polynomial with $e^{ax}$, $\sin x$ or $\ln x$.

Don't do this:

Choosing the order of the sum so that the polynomial ends up as a high-order factor in every term you write, then trying to evaluate $\binom{n}{k}$ for all $k$ up to $n$.

The correct way:

Give the polynomial the low orders $k = 0, 1, 2, \ldots$ and stop when its derivative hits $0$. A quadratic leaves three terms, a cubic four. The rusher who starts writing all $n + 1$ terms before checking which ones vanish usually runs out of time.

Mistake 2: Dropping Or Miscounting The Binomial Coefficients

Where it slips in:

Any order above 1, especially the middle terms.

Don't do this:

Writing $(fg)'' = f''g + f'g' + fg''$, or using $\binom{n}{2} = n - 1$.

The correct way:

Read the coefficients from Pascal's triangle or compute $\binom{n}{2} = \frac{n(n-1)}{2}$. A Physics Forums thread on a "tricky" Leibniz question turns on exactly this: one binomial coefficient worked out as $m - 1$ instead of $m$. The memorizer who learned $1, 2, 1$ for the second derivative but never saw why tends to reuse those coefficients at every order.

Mistake 3: Losing The Sign In Logarithm Derivatives

Where it slips in:

Products such as $x^2\ln x$ and $x^3\ln x$.

Don't do this:

Writing $(\ln x)''' = -\frac{2}{x^3}$.

The correct way:

Use $(\ln x)^{(m)} = (-1)^{m-1}\frac{1\cdot 2\cdots(m-1)}{x^m}$: the signs run $+, -, +, -$ starting from $\frac{1}{x}$. Students who check one value numerically (the sign of $(\ln x)''$ must be negative, because $\ln x$ bends downward) stop flipping these signs.

Mistake 4: Applying The Rule To A Sum Or A Composition

Where it slips in:

Functions like $e^{x^2}$ or $\sin\left(x^2\right)$ that look like products at a glance but are compositions.

Don't do this:

Using the Leibniz rule on $e^{x^2}$ as if it were a product.

The correct way:

The rule is for products only. A composition needs the chain rule (or a recurrence), and a sum is differentiated term by term. The silent understander can run the formula correctly but has never asked what kind of expression it needs; identifying "product of two factors" first prevents this.

Practice Problems On The General Leibniz Rule

Work each problem, then compare with the answers below. Every answer has been recomputed.

1. Find the fourth derivative of $x^2\sin x$.

2. Find the nth derivative of $xe^{3x}$.

3. Find the nth derivative of $e^x(2x + 3)^3$.

4. Find the fifth derivative of $x^3e^{-x}$.

5. Find the third derivative of $x^2\ln x$, for $x > 0$.

6. Find the tenth derivative of $\left(4x^2 + 1\right)e^x$.

Answer to Question 1: $x^2\sin x - 8x\cos x - 12\sin x$.

Answer to Question 2: $3^{n-1}(3x + n)e^{3x}$.

Answer to Question 3: $e^x\left[(2x+3)^3 + 6n(2x+3)^2 + 12n(n-1)(2x+3) + 8n(n-1)(n-2)\right]$.

Answer to Question 4: $-\left(x^3 - 15x^2 + 60x - 60\right)e^{-x}$.

Answer to Question 5: $\frac{2}{x}$.

Answer to Question 6: $\left(4x^2 + 80x + 361\right)e^x$.

Bottom Line

• The general Leibniz rule writes the nth derivative of a product as $\sum_{k=0}^{n}\binom{n}{k}f^{(n-k)}g^{(k)}$.

• Its coefficients are the rows of Pascal's triangle, for the same counting reason as the binomial theorem.

• Induction proves it, with Pascal's rule doing the work in the inductive step.

• Give the polynomial factor the low orders so most terms vanish before you write them.

• Test any expansion on $x \cdot x$ to catch a missing coefficient.

Your Next Move

Work through Practice Problems 1 to 6; if Problem 3 or Problem 6 trips you up, return to the four-step method, and if a sign goes wrong in Problem 5, reread Mistake 3 on the derivatives of $\ln x$.

Book a free demo class with Bhanzu:

Read More

• Second derivative: the order-2 case of repeated differentiation, with its meaning for graphs.

• uv differentiation formula: the first-order product rule written in u and v.

• Binomial expansion formula: the algebra whose coefficients the Leibniz rule reuses.

• Table of derivatives: the base derivatives every Leibniz expansion needs.

• Chain rule: the rule for compositions, which the Leibniz rule does not cover.