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