Select Page

# Calculus

Differentiation of Functions

# Leibniz Formula

Page 1
Problem 1
Page 2
Problems 2-9

The Leibniz formula expresses the derivative on $$n$$th order of the product of two function. Suppose that the functions $$u\left( x \right)$$ and $$v\left( x \right)$$ have the derivatives up to $$n$$th 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 $$uv:$$
${{\left( {uv} \right)^{\prime\prime\prime}} = {\left[ {{\left( {uv} \right)^{\prime\prime}}} \right]^\prime } } = {{\left( {u^{\prime\prime}v + 2u’v’ + uv^{\prime\prime}} \right)^\prime } } = {{\left( {u^{\prime\prime}v} \right)^\prime } + {\left( {2u’v’} \right)^\prime } + {\left( {uv^{\prime\prime}} \right)^\prime } } = {u^{\prime\prime\prime}v + \color{blue}{u^{\prime\prime}v’} + \color{blue}{2u^{\prime\prime}v’} }+{ \color{red}{2u’v^{\prime\prime}} + \color{red}{u’v^{\prime\prime}} + uv^{\prime\prime\prime} } = {u^{\prime\prime\prime}v + \color{blue}{3u^{\prime\prime}v’} }+{ \color{red}{3u’v^{\prime\prime}} + uv^{\prime\prime\prime}.}$ 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
${\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right){u^{\left( {n – m + 1} \right)}}{v^{\left( m \right)}} + \left( {\begin{array}{*{20}{c}} n\\ {m – 1} \end{array}} \right){u^{\left( {n – m + 1} \right)}}{v^{\left( m \right)}} } = {\left[ {\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {m – 1} \end{array}} \right)} \right]\cdot}\kern0pt{{u^{\left( {n – m + 1} \right)}}{v^{\left( m \right)}}.}$ 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
${\left[ {\left( {\begin{array}{*{20}{c}} n\\ m \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {m – 1} \end{array}} \right)} \right]\cdot}\kern0pt{{u^{\left( {n – m + 1} \right)}}{v^{\left( m \right)}} } = {\left( {\begin{array}{*{20}{c}} {n + 1}\\ m \end{array}} \right){u^{\left( {n + 1 – m} \right)}}{v^{\left( m \right)}}.}$ 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 can be seen, 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 on problem description to see solution.

### ✓Example 1

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

### ✓Example 2

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

### ✓Example 3

Given the function $$y = {x^2}\cos 3x.$$ Find the third-order derivative.

### ✓Example 4

Find the fifth derivative of the function $${y }={ \left( {{x^3} + 2{x^2} + 3x} \right){e^x}.}$$

### ✓Example 5

Find the $$n$$th-order derivative of the function $$y = {x^2}\cos x.$$

### ✓Example 6

Find the $$10$$th-order derivative of the function $$y = \left( {{x^2} + 4x + 1} \right)\sqrt {{e^x}}$$ at the point $$x = 0$$.

### ✓Example 7

Find the $$n$$th-order derivative of the function $$y = {x^3}\sin 2x.$$

### ✓Example 8

Find the $$n$$th-order derivative of the function $$y = \left( {3{x^2} – 2x} \right)\ln x,$$ $$x \gt 0.$$

### ✓Example 9

Find the $$n$$th-order derivative of the function $$y = \arctan x$$ at the point $$x = 0.$$

### Example 1.

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
${u’ = {\left( {{e^{2x}}} \right)^\prime } = 2{e^{2x}},\;\;\;}\kern-0.3pt {u^{\prime\prime} = {\left( {2{e^{2x}}} \right)^\prime } = 4{e^{2x}},\;\;\;}\kern-0.3pt {u^{\prime\prime\prime} = {\left( {4{e^{2x}}} \right)^\prime } = 8{e^{2x}},}$ ${v’ = {\left( {\ln x} \right)^\prime } = \frac{1}{x},\;\;\;}\kern-0.3pt {v^{\prime\prime} = {\left( {\frac{1}{x}} \right)^\prime } = – \frac{1}{{{x^2}}},\;\;\;}\kern-0.3pt {v^{\prime\prime\prime} = {\left( { – \frac{1}{{{x^2}}}} \right)^\prime } } = { – {\left( {{x^{ – 2}}} \right)^\prime } } = {2{x^{ – 3}} }={ \frac{2}{{{x^3}}}.}$ The third-order derivative of the original function is given by the Leibniz rule:
${y^{\prime\prime\prime} = {\left( {{e^{2x}}\ln x} \right)^{\prime \prime \prime }} } = {\sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){u^{\left( {3 – i} \right)}}{v^{\left( i \right)}}} } = {\sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){{\left( {{e^{2x}}} \right)}^{\left( {3 – i} \right)}}{{\left( {\ln x} \right)}^{\left( i \right)}}} } = {\left( {\begin{array}{*{20}{c}} 3\\ 0 \end{array}} \right) \cdot 8{e^{2x}}\ln x } + {\left( {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right) \cdot 4{e^{2x}} \cdot \frac{1}{x} } + {\left( {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right) \cdot 2{e^{2x}} \cdot \left( { – \frac{1}{{{x^2}}}} \right) } + {\left( {\begin{array}{*{20}{c}} 3\\ 3 \end{array}} \right){e^{2x}} \cdot \frac{2}{{{x^3}}} } = {1 \cdot 8{e^{2x}}\ln x }+{ 3 \cdot \frac{{4{e^{2x}}}}{x} } – {3 \cdot \frac{{2{e^{2x}}}}{{{x^2}}} }+{ 1 \cdot \frac{{2{e^{2x}}}}{{{x^3}}} } = {8{e^{2x}}\ln x + \frac{{12{e^{2x}}}}{x} }-{ \frac{{6{e^{2x}}}}{{{x^2}}} }+{ \frac{{2{e^{2x}}}}{{{x^3}}} } = {2{e^{2x}}\cdot}\kern0pt{\left( {4\ln x + \frac{6}{x} – \frac{3}{{{x^2}}} + \frac{1}{{{x^3}}}} \right).}$

Page 1
Problem 1
Page 2
Problems 2-9