Calculus

Differentiation of Functions

Differentiation of Functions Logo

Leibniz Formula

  • The Leibniz formula expresses the derivative on \(n\)th order of the product of two functions. 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

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

    AAs 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 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 7

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

    Example 8

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

    Example 9

    Find the \(3\)rd derivative of the function \(y = \large{\frac{{{e^x}}}{x}}\normalsize.\)

    Example 10

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

    Example 11

    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 12

    Find the \(5\)th derivative of the function \(y = {x^2}\sin 2x\) at \(x = 0.\)

    Example 13

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

    Example 14

    Find the \(n\)th order derivative of the function \(y = x\ln x.\)

    Example 15

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

    Example 16

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

    Example 17

    Calculate the \(4\)th derivative of the function \(y = \large{\frac{{{x^2}}}{{x – 1}}}\normalsize\) at \(x = 2.\)

    Example 18

    Determine the \(3\)rd derivative of the function \(y = \large{\frac{{{x^3}}}{{x + 2}}}\normalsize\) at \(x = -1.\)

    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

    \[{x^\prime = 1,\;\;}\kern0pt{x^{\prime\prime} = x^{\prime\prime\prime} \equiv 0.}\]

    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

    \[{y^{\prime\prime\prime} \text{ = }}\kern0pt{1 \cdot \left( { – \cos x} \right) \cdot x + 3 \cdot \left( { – \sin x} \right) \cdot 1 }={ – x\cos x – 3\sin x.}\]

    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

    \[
    {u’ = {\left( {{e^{2x}}} \right)^\prime } = 2{e^{2x}},\;\;\;}\kern-0.3pt
    {u^{\prime\prime} = {\left( {2{e^{2x}}} \right)^\prime } = 4{e^{2x}},\;\;\;}\kern-0.3pt
    {u^{\prime\prime\prime} = {\left( {4{e^{2x}}} \right)^\prime } = 8{e^{2x}},}
    \]

    \[
    {v’ = {\left( {\ln x} \right)^\prime } = \frac{1}{x},\;\;\;}\kern-0.3pt
    {v^{\prime\prime} = {\left( {\frac{1}{x}} \right)^\prime } = – \frac{1}{{{x^2}}},\;\;\;}\kern-0.3pt
    {v^{\prime\prime\prime} = {\left( { – \frac{1}{{{x^2}}}} \right)^\prime } }
    = { – {\left( {{x^{ – 2}}} \right)^\prime } }
    = {2{x^{ – 3}} }={ \frac{2}{{{x^3}}}.}
    \]

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

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

    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

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

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    Problems 1-5
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    Problems 6-18