# Leibniz Formula

The Leibniz formula expresses the derivative on nth order of the product of two functions. Suppose that the functions u (x) and v (x) have the derivatives up to nth order. Consider the derivative of the product of these functions.

The first derivative is described by the well known formula:

${\left( {uv} \right)^\prime } = u'v + uv'.$

Differentiating this expression again yields the second derivative:

$\left( {uv} \right)^{\prime\prime} = \left[ {{{\left( {uv} \right)}^\prime }} \right]^\prime = \left( {u'v + uv'} \right)^\prime = {\left( {u'v} \right)^\prime } + {\left( {uv'} \right)^\prime } = u^{\prime\prime}v + u'v' + u'v' + uv^{\prime\prime} = u^{\prime\prime}v + 2u'v' + uv^{\prime\prime}.$

Likewise, we can find the third derivative of the product $$uv:$$

$\left( {uv} \right)^{\prime\prime\prime} = \left[ {{\left( {uv} \right)^{\prime\prime}}} \right]^\prime = \left( {u^{\prime\prime}v + 2u'v' + uv^{\prime\prime}} \right)^\prime = {\left( {u^{\prime\prime}v} \right)^\prime } + {\left( {2u'v'} \right)^\prime } + {\left( {uv^{\prime\prime}} \right)^\prime } = u^{\prime\prime\prime}v + \color{blue}{u^{\prime\prime}v'} + \color{blue}{2u^{\prime\prime}v'} + \color{red}{2u'v^{\prime\prime}} + \color{red}{u'v^{\prime\prime}} + \color{black}{uv^{\prime\prime\prime}} = u^{\prime\prime\prime}v + \color{blue}{3u^{\prime\prime}v'} + \color{red}{3u'v^{\prime\prime}} + \color{black}{uv^{\prime\prime\prime}}.$

It is easy to see that these formulas are similar to the binomial expansion raised to the appropriate exponent. Assuming that the terms with zero exponent $${u^0}$$ and $${v^0}$$ correspond to the functions $$u$$ and $$v$$ themselves, we can write the general formula for the derivative of $$n$$th order of the product of functions $$uv$$ as follows:

$\left( {uv} \right)^{\left( n \right)} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i} \right)}}{v^{\left( i \right)}}} ,$

where $${\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right)}$$ denotes the number of $$i$$-combinations of $$n$$ elements.

This formula is called the Leibniz formula and can be proved by induction.

### Proof.

Suppose that the functions $$u$$ and $$v$$ have the derivatives of $$\left( {n + 1} \right)$$th order. Using the recurrence relation, we write the expression for the derivative of $$\left( {n + 1} \right)$$th order in the following form:

$y^{\left( {n + 1} \right)} = \left[ {{y^{\left( n \right)}}} \right]^\prime = \left[ {{{\left( {uv} \right)}^{\left( n \right)}}} \right]^\prime = \left[ {\sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i} \right)}}{v^{\left( i \right)}}} } \right]^\prime.$

After differentiation we obtain:

$y^{\left( {n + 1} \right)} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i + 1} \right)}}{v^{\left( i \right)}}} + \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i} \right)}}{v^{\left( {i + 1} \right)}}} .$

Both sums in the right-hand side can be combined into a single sum. Indeed, take an intermediate index $$1 \le m \le n.$$ The first term when $$i = m$$ is written as

$\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right){u^{\left( {n - m + 1} \right)}}{v^{\left( m \right)}},$

and the second term when $$i = m - 1$$ is as follows:

$\left( {\begin{array}{*{20}{c}} n\\ {m - 1} \end{array}} \right){u^{\left( {n - \left( {m - 1} \right)} \right)}}{v^{\left( {\left( {m - 1} \right) + 1} \right)}} = \left( {\begin{array}{*{20}{c}} n\\ {m - 1} \end{array}} \right){u^{\left( {n - m + 1} \right)}}{v^{\left( m \right)}}.$

The sum of these two terms is given by

$\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right){u^{\left( {n - m + 1} \right)}}{v^{\left( m \right)} + \left( {\begin{array}{*{20}{c}} n\\ {m - 1} \end{array}} \right){u^{\left( {n - m + 1} \right)}}{v^{\left( m \right)}} } = \left[ {\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {m - 1} \end{array}} \right)} \right] {u^{\left( {n - m + 1} \right)}}{v^{\left( m \right)}}.$

It is known from combinatorics that

$\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {m - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {n + 1}\\ m \end{array}} \right).$

Therefore, the sum of these two terms can be written as

$\left[ {\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {m - 1} \end{array}} \right)} \right] {u^{\left( {n - m + 1} \right)}}{v^{\left( m \right)}} = \left( {\begin{array}{*{20}{c}} {n + 1}\\ m \end{array}} \right){u^{\left( {n + 1 - m} \right)}}{v^{\left( m \right)}}.$

It is clear that when $$m$$ changes from $$1$$ to $$n$$ this combination will cover all terms of both sums except the term for $$i = 0$$ in the first sum equal to

$\left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){u^{\left( {n - 0 + 1} \right)}}{v^{\left( 0 \right)}} = {u^{\left( {n + 1} \right)}}{v^{\left( 0 \right)}},$

and the term for $$i = n$$ in the second sum equal to

$\left( {\begin{array}{*{20}{c}} n\\ n \end{array}} \right){u^{\left( {n - n} \right)}}{v^{\left( {n + 1} \right)}} = {u^{\left( 0 \right)}}{v^{\left( {n + 1} \right)}}.$

As a result, the derivative of $$\left( {n + 1} \right)$$th order of the product of functions $$uv$$ is represented in the form

$y^{\left( {n + 1} \right)} = {u^{\left( {n + 1} \right)}}{v^{\left( 0 \right)}} + \sum\limits_{m = 1}^n {\left( {\begin{array}{*{20}{c}} {n + 1}\\ m \end{array}} \right){u^{\left( {n + 1 - m} \right)}}{v^{\left( m \right)}}} + {u^{\left( 0 \right)}}{v^{\left( {n + 1} \right)}} = \sum\limits_{m = 0}^{n + 1} {\left( {\begin{array}{*{20}{c}} {n + 1}\\ m \end{array}} \right){u^{\left( {n + 1 - m} \right)}}{v^{\left( m \right)}}}.$

As you can see, the expression for $${y^{\left( {n + 1} \right)}}$$ has a similar form as for the derivative $${y^{\left( n \right)}}.$$ Only now the upper limit of summation is equal to $$n + 1$$ instead of $$n.$$ Thus, the Leibniz formula is proved for an arbitrary natural number $$n.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the $$4$$th derivative of the function $y = {e^x}\sin x.$

### Example 2

Find the $$3$$rd derivative of the function $y = x\sin x.$

### Example 3

Find the third derivative of the function $y = {e^{2x}}\ln x.$

### Example 4

Find the $$3$$rd derivative of the function $y = {e^x}\cos x.$

### Example 5

Find the $$4$$th-order derivative of the function $y = x\sinh x.$

### Example 6

Find all derivatives of the function $y = {e^x}{x^2}.$

### Example 1.

Find the $$4$$th derivative of the function $y = {e^x}\sin x.$

Solution.

Let $$u = \sin x,$$ $$v = {e^x}.$$ Using the Leibniz formula, we can write

$\require{cancel} {y^{\left( 4 \right)} = \left( {{e^x}\sin x} \right)^{\left( 4 \right)}} = \sum\limits_{i = 0}^4 {\left( {\begin{array}{*{20}{c}} 4\\ i \end{array}} \right){u^{\left( {4 - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^4 {\left( {\begin{array}{*{20}{c}} 4\\ i \end{array}} \right){{\left( {\sin x} \right)}^{\left( {4 - i} \right)}}{{\left( {{e^x}} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} 4\\ 0 \end{array}} \right){\left( {\sin x} \right)^{\left( 4 \right)}}{e^x} + \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right)\left( {\sin x} \right)^{\prime\prime\prime}\left( {{e^x}} \right)^\prime + \left( {\begin{array}{*{20}{c}} 4\\ 2 \end{array}} \right)\left( {\sin x} \right)^{\prime\prime}\left( {{e^x}} \right)^{\prime\prime} + \left( {\begin{array}{*{20}{c}} 4\\ 3 \end{array}} \right)\left( {\sin x} \right)^\prime\left( {{e^x}} \right)^{\prime\prime\prime} + \left( {\begin{array}{*{20}{c}} 4\\ 4 \end{array}} \right)\left( {\sin x} \right){\left( {{e^x}} \right)^{\left( 4 \right)}} = 1 \cdot \sin x \cdot {e^x} + \cancel{ 4 \cdot \left( { - \cos x} \right) \cdot {e^x} } + 6 \cdot \left( { - \sin x} \right) \cdot {e^x} + \cancel{ 4 \cdot \cos x \cdot {e^x} } + 1 \cdot \sin x \cdot {e^x} = - 4{e^x}\sin x.$

### Example 2.

Find the $$3$$rd derivative of the function $y = x\sin x.$

Solution.

Let $$u = \sin x,$$ $$v = x.$$ By the Leibniz formula, we can write:

$y^{\prime\prime\prime} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){u^{\left( {3 - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){{\left( {\sin x} \right)}^{\left( {3 - i} \right)}}{x^{\left( i \right)}}} .$

It is clear that

$x^\prime = 1,\;\; x^{\prime\prime} = x^{\prime\prime\prime} \equiv 0.$

Then the series expansion has only two terms:

$y^{\prime\prime\prime} = \left( {\begin{array}{*{20}{c}} 3\\ 0 \end{array}} \right)\left( {\sin x} \right)^{\prime\prime\prime}x + \left( {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right)\left( {\sin x} \right)^{\prime\prime}x^\prime.$

Calculating the derivatives, we obtain

$y^{\prime\prime\prime} = 1 \cdot \left( { - \cos x} \right) \cdot x + 3 \cdot \left( { - \sin x} \right) \cdot 1 = - x\cos x - 3\sin x.$

### Example 3.

Find the third derivative of the function $y = {e^{2x}}\ln x.$

Solution.

We set $$u = {e^{2x}}$$, $$v = \ln x$$. The derivatives of the functions $$u$$ and $$v$$ are

$u' = \left( {{e^{2x}}} \right)^\prime = 2{e^{2x}},\;\;\;u^{\prime\prime} = {\left( {2{e^{2x}}} \right)^\prime } = 4{e^{2x}},\;\;\;u^{\prime\prime\prime} = {\left( {4{e^{2x}}} \right)^\prime } = 8{e^{2x}},$
$v' = \left( {\ln x} \right)^\prime = \frac{1}{x},\;\;\;v^{\prime\prime} = {\left( {\frac{1}{x}} \right)^\prime } = - \frac{1}{{{x^2}}},\;\;\;v^{\prime\prime\prime} = {\left( { - \frac{1}{{{x^2}}}} \right)^\prime } = - {\left( {{x^{ - 2}}} \right)^\prime } = 2{x^{ - 3}} = \frac{2}{{{x^3}}}.$

The third-order derivative of the original function is given by the Leibniz rule:

${y^{\prime\prime\prime} = {\left( {{e^{2x}}\ln x} \right)^{\prime \prime \prime }} } = {\sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){u^{\left( {3 - i} \right)}}{v^{\left( i \right)}}} } = {\sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){{\left( {{e^{2x}}} \right)}^{\left( {3 - i} \right)}}{{\left( {\ln x} \right)}^{\left( i \right)}}} } = {\left( {\begin{array}{*{20}{c}} 3\\ 0 \end{array}} \right) \cdot 8{e^{2x}}\ln x } + {\left( {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right) \cdot 4{e^{2x}} \cdot \frac{1}{x} } + {\left( {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right) \cdot 2{e^{2x}} \cdot \left( { - \frac{1}{{{x^2}}}} \right) } + {\left( {\begin{array}{*{20}{c}} 3\\ 3 \end{array}} \right){e^{2x}} \cdot \frac{2}{{{x^3}}} } = {1 \cdot 8{e^{2x}}\ln x }+{ 3 \cdot \frac{{4{e^{2x}}}}{x} } - {3 \cdot \frac{{2{e^{2x}}}}{{{x^2}}} }+{ 1 \cdot \frac{{2{e^{2x}}}}{{{x^3}}} } = {8{e^{2x}}\ln x + \frac{{12{e^{2x}}}}{x} }-{ \frac{{6{e^{2x}}}}{{{x^2}}} }+{ \frac{{2{e^{2x}}}}{{{x^3}}} } = {2{e^{2x}}\cdot}\kern0pt{\left( {4\ln x + \frac{6}{x} - \frac{3}{{{x^2}}} + \frac{1}{{{x^3}}}} \right).}$

### Example 4.

Find the $$3$$rd derivative of the function $y = {e^x}\cos x.$

Solution.

Let $$u = \cos x,$$ $$v = {e^x}.$$ Using the Leibniz formula, we have

$y^{\prime\prime\prime} = \left( {{e^x}\cos x} \right)^{\prime\prime\prime} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){{\left( {\cos x} \right)}^{\left( {3 - i} \right)}}{{\left( {{e^x}} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} 3\\ 0 \end{array}} \right)\left( {\cos x} \right)^{\prime\prime\prime}{e^x} + \left( {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right)\left( {\cos x} \right)^{\prime\prime}\left( {{e^x}} \right)^\prime + \left( {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right)\left( {\cos x} \right)^\prime\left( {{e^x}} \right)^{\prime\prime} + \left( {\begin{array}{*{20}{c}} 3\\ 3 \end{array}} \right)\cos x\left( {{e^x}} \right)^{\prime\prime\prime}.$

The derivatives of cosine are

$\left( {\cos x} \right)^\prime = - \sin x;$
$\left( {\cos x} \right)^{\prime\prime} = \left( { - \sin x} \right)\prime = - \cos x;$
$\left( {\cos x} \right)^{\prime\prime\prime} = \left( { - \cos x} \right)\prime = \sin x.$

All derivatives of the exponential function $$v = {e^x}$$ are $${e^x}.$$ Hence,

$y^{\prime\prime\prime} = 1 \cdot \sin x \cdot {e^x} + 3 \cdot \left( { - \cos x} \right) \cdot {e^x} + 3 \cdot \left( { - \sin x} \right) \cdot {e^x} + 1 \cdot \cos x \cdot {e^x} = {e^x}\left( { - 2\sin x - 2\cos x} \right) = - 2{e^x}\left( {\sin x + \cos x} \right).$

### Example 5.

Find the $$4$$th-order derivative of the function $y = x\sinh x.$

Solution.

We denote $$u = \sinh x,$$ $$v = x.$$ By the Leibniz formula,

${y^{\left( 4 \right)}} = {\left( {x\sinh x} \right)^{\left( 4 \right)}} = \sum\limits_{i = 0}^4 {\left( {\begin{array}{*{20}{c}} 4\\ i \end{array}} \right){{\left( {\sinh x} \right)}^{\left( {4 - i} \right)}}{x^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} 4\\ 0 \end{array}} \right){\left( {\sinh x} \right)^{\left( 4 \right)}}x + \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right){\left( {\sinh x} \right)^{\left( 3 \right)}}x^\prime + \ldots$

Calculate the derivatives of the hyperbolic sine function:

$\left( {\sinh } \right)^\prime = \cosh x;$
$\left( {\sinh } \right)^{\prime\prime} = \left( {\cosh x} \right)^\prime = \sinh x;$
$\left( {\sinh } \right)^{\prime\prime\prime} = \left( {\sinh x} \right)^\prime = \cosh x;$
$\left( {\sinh } \right)^{\left( 4 \right)} = \left( {\cosh x} \right)^\prime = \sinh x.$

So, this yields

${y^{\left( 4 \right)}} = \left( {\begin{array}{*{20}{c}} 4\\ 0 \end{array}} \right)\sinh x \cdot x + \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right)\cosh x \cdot 1 = 1 \cdot \sinh x \cdot x + 4 \cdot \cosh x \cdot 1 = x\sinh x + 4\cosh x.$

### Example 6.

Find all derivatives of the function $y = {e^x}{x^2}.$

Solution.

Let $$u = {e^x}$$ and $$v = {x^2}$$. Then

$u' = {\left( {{e^x}} \right)^\prime } = {e^x},\;\;\;v' = {\left( {{x^2}} \right)^\prime } = 2x,\;\;\;u^{\prime\prime} = {\left( {{e^x}} \right)^\prime } = {e^x},\;\;\;v^{\prime\prime} = {\left( {2x} \right)^\prime } = 2.$

It is easy to find the general formulas for the derivatives of order $$n:$$

${u^{\left( n \right)}} = {e^x},\;\;\;v^{\prime\prime\prime} = {v^{IV}} = \ldots = {v^{\left( n \right)}} = 0.$

Using the Leibniz formula

$\left( {uv} \right)^{\left( n \right) = u^{\left( n \right)}}v + n{u^{\left( {n - 1} \right)}}v' + \frac{{n\left( {n - 1} \right)}}{{1 \cdot 2}}{u^{\left( {n - 2} \right)}}v^{\prime\prime} + \ldots + u{v^{\left( n \right)}},$

we obtain

${y^{\left( n \right)}} = {e^x}{x^2} + n{e^x} \cdot 2x + \frac{{n\left( {n - 1} \right)}}{{1 \cdot 2}}{e^x} \cdot 2\;\;\;\text{or}\;\;\;{y^{\left( n \right)}} = {e^x} \left[ {{x^2} + 2nx + n\left( {n - 1} \right)} \right].$

See more problems on Page 2.