Calculus

Differentiation of Functions

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Leibniz Formula

Solved Problems

Click or tap a problem to see the solution.

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 \[y = \frac{{{e^x}}}{x}.\]

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 \[y = \left( {3{x^2} - 2x} \right)\ln x,\] where \(x \gt 0.\)

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 \[y = \frac{{{x^2}}}{{x - 1}}\] at \(x = 2.\)

Example 18

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

Example 7.

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

Solution.

Let \(u = \cos 3x\), \(v = {x^2}\). Then by the Leibniz formula, we find:

\[y^{\prime\prime\prime} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){u^{\left( {3 - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){{\left( {\cos 3x} \right)}^{\left( {3 - i} \right)}}{{\left( {{x^2}} \right)}^{\left( i \right)}}} .\]

The derivatives in this expression are of the form:

\[{\left( {\cos 3x} \right)'} = - 3\sin 3x,\;\;\;\left( {\cos 3x} \right)^{\prime\prime} = \left( { - 3\sin 3x} \right)^\prime = - 9\cos 3x,\;\;\;\left( {\cos 3x} \right)^{\prime\prime\prime} = \left( { - 9\cos 3x} \right)^\prime = 27\sin 3x,\]
\[\left( {{x^2}} \right)^\prime = 2x,\;\;\;\left( {{x^2}} \right)^{\prime\prime} = 2,\;\;\;\left( {{x^2}} \right)^{\prime\prime\prime} = 0.\]

Consequently, the third derivative of the given function is

\[ y^{\prime\prime\prime} = \left( {\begin{array}{*{20}{c}} 3\\ 0 \end{array}} \right){\left( {\cos 3x} \right)^{\prime\prime\prime}}{x^2} + \left( {\begin{array}{*{20}{c}} 3\\ 1 \end{array}} \right){\left( {\cos 3x} \right)^{\prime\prime} } {\left( {{x^2}} \right)^\prime } + \left( {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right){\left( {\cos 3x} \right)^\prime }{\left( {{x^2}} \right)^{\prime\prime} } + \left( {\begin{array}{*{20}{c}} 3\\ 3 \end{array}} \right)\cos 3x{\left( {{x^2}} \right)^{\prime\prime\prime}} = 1 \cdot 27\sin 3x \cdot {x^2} + 3 \cdot \left( { - 9\cos 3x} \right) \cdot 2x + 3 \cdot \left( { - 3\sin 3x} \right) \cdot 2 + 1 \cdot \cos 3x \cdot 0 = 27{x^2}\sin 3x - 54x\cos 3x - 18\sin 3x = \left( {27{x^2} - 18} \right)\sin 3x - 54x\cos 3x. \]

Example 8.

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

Solution.

To calculate the derivative \({y^{\left( 5 \right)}}\) we apply the Leibniz rule. Let \(v = {x^3} + 2{x^2} + 3x\), \(u = {e^x}.\) Then the \(5\)th-order derivative \({y^{\left( 5 \right)}}\) is represented as the following series:

\[y^{\left( 5 \right)} = \sum\limits_{i = 0}^5 {\left( {\begin{array}{*{20}{c}} 5\\ i \end{array}} \right){u^{\left( {5 - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^5 {\left( {\begin{array}{*{20}{c}} 5\\ i \end{array}} \right){{\left( {{e^x}} \right)}^{\left( {5 - i} \right)}} {{\left( {{x^3} + 2{x^2} + 3x} \right)}^{\left( i \right)}}} . \]

The derivative of any order of the exponential function \(u = {e^x}\) is the exponential function itself:

\[u^{\left( {5 - i} \right)} = \left( {{e^x}} \right)^{\left( {5 - i} \right)} \equiv e^x,\]

and the derivative of the polynomial \(v = {x^3} + 2{x^2} + 3x\) is nonzero only for the first three orders of differentiation:

\[v' = \left( {{x^3} + 2{x^2} + 3x} \right)^\prime = 3{x^2} + 4x + 3,\;\;\;v^{\prime\prime} = \left( {3{x^2} + 4x + 3} \right)^\prime = {6x + 4,}\;\;\;v^{\prime\prime\prime} = \left( {6x + 4} \right)^\prime = 6,\;\;\;v^{\left( 4 \right)} = v^{\left( 5 \right)} \equiv 0.\]

Hence, the series expansion of the derivative \({y^{\left( 5 \right)}}\) has the form

\[y^{\left( 5 \right)} = \left( {\begin{array}{*{20}{c}} 5\\ 0 \end{array}} \right){e^x}\left( {{x^3} + 2{x^2} + 3x} \right) + \left( {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right){e^x}\left( {3{x^2} + 4x + 3} \right) + \left( {\begin{array}{*{20}{c}} 5\\ 2 \end{array}} \right){e^x}\left( {6x + 4} \right) + \left( {\begin{array}{*{20}{c}} 5\\ 3 \end{array}} \right){e^x} \cdot 6 \]

The remaining terms of the series are obviously equal to zero. As a result, we have

\[y^{\left( 5 \right)} = {e^x}\left( {{x^3} + 2{x^2} + 3x} \right) + 5{e^x}\left( {3{x^2} + 4x + 3} \right) + 10{e^x}\left( {6x + 4} \right) + 60{e^x} = {e^x}\left( {{x^3} + {17{x^2}} + {83x} + {115}} \right).\]

Example 9.

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

Solution.

Let \(u = {e^x},\) \(v = \frac{1}{x}.\)

All derivatives of the exponential function \(u = {e^x}\) are \({e^x}:\)

\[u^\prime = u^{\prime\prime} = u^{\prime\prime\prime} = \ldots = {u^{\left( n \right)}} = {e^x}.\]

Compute the first derivatives of \(v = \frac{1}{x}:\)

\[v^\prime = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\;\;v^{\prime\prime} = \left( { - \frac{1}{{{x^2}}}} \right)^\prime = \frac{2}{{{x^3}}},\;\;v^{\prime\prime\prime} = \left( {\frac{2}{{{x^3}}}} \right)^\prime = - \frac{6}{{{x^4}}}.\]

The common pattern is clear, so

\[v^{\left( n \right)} = \left( {\frac{1}{x}} \right)^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^n}n!}}{{{x^{n + 1}}}}.\]

We apply the Leibniz rule:

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

Substitute the expressions for the derivatives:

\[y^{\prime\prime\prime} = 1 \cdot {e^x} \cdot \frac{1}{x} + 3 \cdot {e^x} \cdot \left( { - \frac{1}{{{x^2}}}} \right) + 3 \cdot {e^x} \cdot \frac{2}{{{x^3}}} + 1 \cdot {e^x} \cdot \left( { - \frac{6}{{{x^4}}}} \right) = {e^x}\left( {\frac{1}{x} - \frac{3}{{{x^2}}} + \frac{6}{{{x^3}}} - \frac{6}{{{x^4}}}} \right) = \frac{{{e^x}}}{{{x^4}}}\left( {{x^3} - 3{x^2} + 6x - 6} \right).\]

Example 10.

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

Solution.

We use the Leibniz formula assuming \(u = \cos x\), \(v = {x^2}.\) Then

\[\left( {{x^2}\cos x} \right)^{\left( n \right)} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){{\left( {\cos x} \right)}^{\left( {n - i} \right)}}{{\left( {{x^2}} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){\left( {\cos x} \right)^{\left( n \right)}}{x^2} + \left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right){\left( {\cos x} \right)^{\left( {n - 1} \right)}}{\left( {{x^2}} \right)^\prime } + \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right){\left( {\cos x} \right)^{\left( {n - 2} \right)}}{\left( {{x^2}} \right)^{\prime\prime}} + \ldots \]

The other terms of the series are zero since \({\left( {{x^2}} \right)^{\left( i \right)}} = 0\) for \(i \gt 2.\)

The \(n\)th-order derivative of the cosine function was determined on the Higher-Order Derivatives page:

\[{\left( {\cos x} \right)^{\left( n \right)}} = \cos \left( {x + \frac{{\pi n}}{2}} \right).\]

Consequently, the derivative of our function is given by

\[{\left( {{x^2}\cos x} \right)^{\left( n \right)}} = C_n^0\cos \left( {x + \frac{{\pi n}}{2}} \right){x^2} + C_n^1\cos \left( {x + \frac{{\pi \left( {n - 1} \right)}}{2}} \right) \cdot 2x + C_n^2\cos \left( {x + \frac{{\pi \left( {n - 2} \right)}}{2}} \right) \cdot 2 = {x^2}\cos \left( {x + \frac{{\pi n}}{2}} \right) + 2nx\cos \left( {x + \frac{{\pi n}}{2} - \frac{\pi }{2}} \right) + \frac{{2n\left( {n - 1} \right)}}{2}\cos \left( {x + \frac{{\pi n}}{2} - \pi } \right) = {x^2}\cos \left( {x + \frac{{\pi n}}{2}} \right) + 2nx\sin \left( {x + \frac{{\pi n}}{2}} \right) - n\left( {n - 1} \right)\cos \left( {x + \frac{{\pi n}}{2}} \right) = \left[ {{x^2} - n\left( {n - 1} \right)} \right] \cos \left( {x + \frac{{\pi n}}{2}} \right) + 2nx\sin \left( {x + \frac{{\pi n}}{2}} \right). \]

Example 11.

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

Solution.

We denote \(u = \sqrt {{e^x}} \), \(v = {x^2} + 4x + 1.\) The derivatives of these functions have the following form:

\[u' = \left( {\sqrt {{e^x}} } \right)^\prime = \frac{1}{{2\sqrt {{e^x}} }} \cdot {\left( {{e^x}} \right)^\prime = \frac{{{e^x}}}{{2\sqrt {{e^x}} }} = \frac{{\sqrt {{e^x}} }}{2},}\;\;\; u^{\prime\prime = \left( {\frac{{\sqrt {{e^x}} }}{2}} \right)^\prime = \frac{{\sqrt {{e^x}} }}{4}, \ldots} \; \Rightarrow u^{\left( k \right)} = \frac{{\sqrt {{e^x}} }}{{{2^k}}},\]
\[v' = {\left( {{x^2} + 4x + 1} \right)^\prime } = {2x + 4,}\;\;\;v^{\prime\prime} = {\left( {2x + 4} \right)^\prime } = 2.\]

The derivatives of the function \(v\) of order \(i \gt 2\) are obviously zero. Therefore, the expansion of the derivative \({y^{\left( {10} \right)}}\) is limited to only a few terms:

\[y^{\left( {10} \right)} = \sum\limits_{i = 0}^{10} {\left( {\begin{array}{*{20}{c}} 10\\ i \end{array}} \right){u^{\left( {10 - i} \right)}}{v^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} 10\\ 0 \end{array}} \right)\frac{{\sqrt {{e^x}} }}{{{2^{10}}}}\left( {{x^2} + 4x + 1} \right) + \left( {\begin{array}{*{20}{c}} 10\\ 1 \end{array}} \right)\frac{{\sqrt {{e^x}} }}{{{2^9}}}\left( {2x + 4} \right) + \left( {\begin{array}{*{20}{c}} 10\\ 2 \end{array}} \right)\frac{{\sqrt {{e^x}} }}{{{2^8}}} \cdot 2 = \frac{{10!}}{{10!\;0!}} \cdot \sqrt {{e^x}} \cdot \frac{1}{{{2^{10}}}} \cdot \left( {{x^2} + 4x + 1} \right) + \frac{{10!}}{{9!\;1!}} \cdot \sqrt {{e^x}} \cdot \frac{2}{{{2^{10}}}} \cdot \left( {2x + 4} \right) + \frac{{10!}}{{8!\;2!}} \cdot \sqrt {{e^x}} \cdot \frac{4}{{{2^{10}}}} \cdot 2 = \frac{{\sqrt {{e^x}} }}{{{2^{10}}}}\cdot \left[ {{x^2} + 4x + 1 + 20\left( {2x + 4} \right) + 360} \right] = \frac{{\sqrt {{e^x}} }}{{{2^{10}}}}\left( {{x^2} + 44x + 441} \right). \]

When \(x = 0\), the \(10\)th-order derivative is respectively equal to

\[{y^{\left( {10} \right)}}\left( 0 \right) = \frac{{441}}{{{2^{10}}}} = \frac{{441}}{{1024}} = {\left( {\frac{{21}}{{32}}} \right)^2}.\]

Example 12.

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

Solution.

We apply the Leibniz formula

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

Let \(u = \sin 2x\) and \(v = {x^2}.\) Then we can write

\[y^{\left( 5 \right)} = \left( {{x^2}\sin 2x} \right)^{\left( 5 \right)} = \sum\limits_{i = 0}^5 {\left( {\begin{array}{*{20}{c}} 5\\ i \end{array}} \right){{\left( {\sin 2x} \right)}^{\left( {5 - i} \right)}}{{\left( {{x^2}} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} 5\\ 0 \end{array}} \right){\left( {\sin 2x} \right)^{\left( 5 \right)}}{x^2} + \left( {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right){\left( {\sin 2x} \right)^{\left( 4 \right)}}\left( {{x^2}} \right)^\prime + \left( {\begin{array}{*{20}{c}} 5\\ 2 \end{array}} \right)\left( {\sin 2x} \right)^{\prime\prime\prime}\left( {{x^2}} \right)^{\prime\prime} + \ldots \]

The remaining terms in the series expansion are zero as \({\left( {{x^2}} \right)^{\left( i \right)}} \equiv 0\) for \(i \gt 2.\)

Determine the derivatives of the sine function. It is known that

\[\left( {\sin x} \right)^{\left( n \right)} = \sin \left( {x + \frac{{\pi n}}{2}} \right).\]

One can show that

\[\left( {\sin 2x} \right)^{\left( n \right)} = {2^n}\sin \left( {2x + \frac{{\pi n}}{2}} \right).\]

Hence

\[\left( {\sin 2x} \right)^{\left( 5 \right)} = {2^5}\sin \left( {2x + \frac{{5\pi }}{2}} \right) = {2^5}\sin \left( {2x + 2\pi + \frac{\pi }{2}} \right) = {2^5}\sin \left( {2x + \frac{\pi }{2}} \right) = {2^5}\cos 2x;\]
\[\left( {\sin 2x} \right)^{\left( 4 \right)} = {2^4}\sin \left( {2x + \frac{{4\pi }}{2}} \right) = {2^4}\sin \left( {2x + 2\pi } \right) = {2^4}\sin 2x;\]
\[\left( {\sin 2x} \right)^{\prime\prime\prime} = {2^3}\sin \left( {2x + \frac{{3\pi }}{2}} \right) = - {2^3}\cos 2x.\]

So the fifth-order derivative is given by

\[y^{\left( 5 \right)} = \left( {\begin{array}{*{20}{c}} 5\\ 0 \end{array}} \right) \cdot {2^5}\cos 2x \cdot {x^2} + \left( {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right) \cdot {2^4}\sin 2x \cdot 2x - \left( {\begin{array}{*{20}{c}} 5\\ 2 \end{array}} \right) \cdot {2^3}\cos 2x \cdot 2 = 1 \cdot {2^5}\cos 2x \cdot {x^2} + 5 \cdot {2^4}\sin 2x \cdot 2x - 10 \cdot {2^3}\cos 2x \cdot 2 = 32{x^2}\cos 2x + 160x\sin 2x - 160\cos 2x = \left( {32{x^2} - 160} \right)\cos 2x + 160x\sin 2x.\]

Substituting \(x = 0,\) we get

\[y^{\left( 5 \right)}\left( 0 \right) = - 160\cos 0 = - 160.\]

Example 13.

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

Solution.

Let \(u = \sin 2x\), \(v = {x^3}.\) Write the \(n\)th-order derivative by the Leibniz formula:

\[ \left( {{x^3}\sin 2x} \right)^{\left( n \right)} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){{\left( {\sin 2x} \right)}^{\left( {n - i} \right)}}{{\left( {{x^3}} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){\left( {\sin 2x} \right)^{\left( n \right)}}{x^3} + \left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right){\left( {\sin 2x} \right)^{\left( {n - 1} \right)}}{\left( {{x^3}} \right)'} + \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right){\left( {\sin 2x} \right)^{\left( {n - 2} \right)}}{\left( {{x^3}} \right)^{\prime\prime}} + \left( {\begin{array}{*{20}{c}} n\\ 3 \end{array}} \right){\left( {\sin 2x} \right)^{\left( {n - 3} \right)}}{\left( {{x^3}} \right)^{\prime\prime\prime}} + \ldots \]

Obviously, the remaining terms in the series expansion are zero since \({\left( {{x^3}} \right)^{\left( i \right)}} = 0\) for \(i \gt 3.\)

The \(n\)th-order derivative of the sine function was found on the Higher-Order Derivatives page. It is written in the form

\[\left( {\sin x} \right)^{\left( n \right)} = \sin \left( {x + \frac{{\pi n}}{2}} \right).\]

It can be shown that the derivative of \({\sin 2x}\) is defined by the similar formula:

\[\left( {\sin 2x} \right)^{\left( n \right)} = {2^n}\sin \left( {2x + \frac{{\pi n}}{2}} \right).\]

Consequently, the remaining derivatives of \({\sin 2x}\) are given by

\[{\left( {\sin 2x} \right)^{\left( {n - 1} \right)}} = {2^{n - 1}}\sin \left( {2x + \frac{{\pi \left( {n - 1} \right)}}{2}} \right) = {2^{n - 1}}\sin \left( {2x + \frac{{\pi n}}{2} - \frac{\pi }{2}} \right) = - {2^{n - 1}}\cos\left( {2x + \frac{{\pi n}}{2}} \right),\]
\[{\left( {\sin 2x} \right)^{\left( {n - 2} \right)}} = {2^{n - 2}}\sin \left( {2x + \frac{{\pi \left( {n - 2} \right)}}{2}} \right) = {2^{n - 2}}\sin \left( {2x + \frac{{\pi n}}{2} - \pi } \right) = - {2^{n - 2}}\sin\left( {2x + \frac{{\pi n}}{2}} \right),\]
\[{\left( {\sin 2x} \right)^{\left( {n - 3} \right)}} = {2^{n - 3}}\sin \left( {2x + \frac{{\pi \left( {n - 3} \right)}}{2}} \right) = {2^{n - 3}}\sin \left( {2x + \frac{{\pi n}}{2} - \frac{{3\pi }}{2}} \right) = {2^{n - 3}}\cos\left( {2x + \frac{{\pi n}}{2}} \right).\]

Substituting this into the formula for the \(n\)th derivative of the given function, we obtain:

\[ \left( {{x^3}\sin 2x} \right)^{\left( n \right)} = \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){x^3}{2^n}\sin\left( {2x + \frac{{\pi n}}{2}} \right) - \left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right) \cdot 3{x^2}{2^{n - 1}}\cos\left( {2x + \frac{{\pi n}}{2}} \right) - \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) \cdot 6x \,{2^{n - 2}}\sin\left( {2x + \frac{{\pi n}}{2}} \right) + \left( {\begin{array}{*{20}{c}} n\\ 3 \end{array}} \right) \cdot 6 \cdot {2^{n - 3}}\cos\left( {2x + \frac{{\pi n}}{2}} \right).\]

Take into account that the combinations can be represented in the following form:

\[ \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right) = 1,\;\;\left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right) = n,\;\;\left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right) = \frac{{n\left( {n - 1} \right)}}{2},\;\;\left( {\begin{array}{*{20}{c}} n\\ 3 \end{array}} \right) = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6}.\]

Then

\[\left( {{x^3}\sin 2x} \right)^{\left( n \right)} = {x^3}{2^n}\sin\left( {2x + \frac{{\pi n}}{2}} \right) - 3{x^2}n{2^{n - 1}}\cos\left( {2x + \frac{{\pi n}}{2}} \right) - 6x \cdot \frac{{n\left( {n - 1} \right)}}{2} \cdot {2^{n - 2}}\sin\left( {2x + \frac{{\pi n}}{2}} \right) + 6 \cdot \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6}\cdot {2^{n - 3}}\cos\left( {2x + \frac{{\pi n}}{2}} \right) = {2^n}\left[ {{x^3} - \frac{{3xn\left( {n - 1} \right)}}{4}} \right] \sin\left( {2x + \frac{{\pi n}}{2}} \right) + {2^n}\left[ {\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{8} - \frac{{3{x^2}n}}{2}} \right] \cos\left( {2x + \frac{{\pi n}}{2}} \right).\]

Example 14.

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

Solution.

Let \(u = \ln x,\) \(v = x.\) Then

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

The other terms of the series are equal to zero as \({x^{\left( i \right)}} \equiv 0\) for \(i \gt 1.\)

Write the derivatives of \(v = x:\)

\[v^\prime = x^\prime = 1,\;\;v^{\prime\prime} = v^{\prime\prime\prime} = \ldots = v^{\left( n \right)} \equiv 0.\]

Compute the derivatives of \(u = \ln x:\)

\[u^\prime = \left( {\ln x} \right)^\prime = \frac{1}{x},\;\;u^{\prime\prime} = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\;\;u^{\prime\prime\prime} = \left( { - \frac{1}{{{x^2}}}} \right)^\prime = \frac{2}{{{x^3}}},\;\;{u^{\left( 4 \right)}} = \left( {\frac{2}{{{x^3}}}} \right)^\prime = - \frac{6}{{{x^4}}},\;\; \ldots \]

So, the \(n\)th derivative of the natural logarithm is written in the form

\[u^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}}.\]

Hence, the series expansion for \({y^{\left( n \right)}}\) is given by

\[{y^{\left( n \right)}} = \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){\left( {\ln x} \right)^{\left( n \right)}} \cdot x + \left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right){\left( {\ln x} \right)^{\left( {n - 1} \right)}} \cdot 1 = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}} \cdot x + n \cdot \frac{{{{\left( { - 1} \right)}^{n - 2}}\left( {n - 2} \right)!}}{{{x^{n - 1}}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}n!}}{{n{x^{n - 1}}}} + \frac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( { - 1} \right)}^{ - 1}}n!}}{{\left( {n - 1} \right){x^{n - 1}}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}n!}}{{{x^{n - 1}}}}\left( {\frac{1}{n} - \frac{1}{{n - 1}}} \right) = \frac{{{{\left( { - 1} \right)}^{n - 1}}n!}}{{{x^{n - 1}}}} \cdot \frac{{\cancel{n} - 1 - \cancel{n}}}{{n\left( {n - 1} \right)}} = \frac{{{{\left( { - 1} \right)}^n}\left( {n - 2} \right)!\cancel{\left( {n - 1} \right)}\cancel{n}}}{{{x^{n - 1}}\cancel{n}\cancel{\left( {n - 1} \right)}}} = \frac{{{{\left( { - 1} \right)}^n}\left( {n - 2} \right)!}}{{{x^{n - 1}}}}.\]

Example 15.

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

Solution.

The derivative of this function can be found by the Leibniz rule:

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

We denote \(u = \ln x\), \(v = 3{x^2} - 2x.\) Write the derivatives of the first function \(u = \ln x:\)

\[u' = \left( {\ln x} \right)^\prime = \frac{1}{x},\;\;\; u^{\prime\prime} = \left( {\frac{1}{x}} \right)^\prime = \left( {{x^{ - 1}}} \right)^\prime = - 1 \cdot {x^{ - 2}} = - \frac{1}{{{x^2}}},\;\;\; u^{\prime\prime\prime} = \left( { - 1 \cdot {x^{ - 2}}} \right)^\prime = \left( { - 1} \right)\left( { - 2} \right){x^{ - 3}} = \frac{2}{{{x^3}}}.\]

It is clear that the \(n\)th-order derivative of the natural logarithm is given by the formula

\[u^{\left( n \right)} = \left( {\ln x} \right)^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}},\]

which can be derived by induction. Regarding the quadratic function \(v = 3{x^2} - 2x,\) it only has non-zero derivatives of the first and second order:

\[v' = \left( {3{x^2} - 2x} \right)^\prime = 6x - 2,\;\;\; v^{\prime\prime} = 6,\;\;\; v^{\prime\prime\prime} = v^{\left( 4 \right)} = \ldots = v^{\left( n \right)} \equiv 0.\]

Therefore, the series expansion of the derivative by the Leibniz rule contains only a few terms:

\[y^{\left( n \right)} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){u^{\left( {n - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ i \end{array}} \right){{\left( {\ln x} \right)}^{\left( {n - i} \right)}} {{\left( {3{x^2} - 2x} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){\left( {\ln x} \right)^{\left( n \right)}} \cdot \left( {3{x^2} - 2x} \right) + \left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right){\left( {\ln x} \right)^{\left( {n - 1} \right)}} \cdot \left( {6x - 2} \right) + \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right){\left( {\ln x} \right)^{\left( {n - 1} \right)}} \cdot 6 = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}} \cdot \left( {3{x^2} - 2x} \right) + n \cdot \frac{{{{\left( { - 1} \right)}^{n - 2}}\left( {n - 2} \right)!}}{{{x^{n - 1}}}} \cdot \left( {6x - 2} \right) + \frac{{n\left( {n - 1} \right)}}{2} \cdot \frac{{{{\left( { - 1} \right)}^{n - 3}}\left( {n - 3} \right)!}}{{{x^{n - 2}}}} \cdot 6 = \frac{{{{\left( { - 1} \right)}^{n - 1}}n!}}{{n{x^{n - 1}}}} \cdot \left( {3x - 2} \right) + \frac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( { - 1} \right)}^{ - 1}}n!}}{{\left( {n - 1} \right){x^{n - 1}}}} \cdot \left( {6x - 2} \right) + \frac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( { - 1} \right)}^{ - 2}}n!x}}{{\left( {n - 2} \right){x^{n - 1}}}} \cdot 3 = {\frac{{{{\left( { - 1} \right)}^{n - 1}}n!}}{{{x^{n - 1}}}}} {\left( {\frac{{3x - 2}}{n} - \frac{{6x - 2}}{{n - 1}} + \frac{{3x}}{{n - 2}}} \right).} \]

This formula is valid for \(n \gt 2.\)

Example 16.

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

Solution.

The first derivative of the inverse tangent function is given by

\[y' = \left( {\arctan x} \right)^\prime = \frac{1}{{1 + {x^2}}}.\]

This expression can be written as

\[y'\left( {1 + {x^2}} \right) = 1.\]

Using the Leibniz rule, we find the derivative of the \(\left( {n - 1} \right)\)th order of both sides of the last equation:

\[\left[ {y'\left( {1 + {x^2}} \right)} \right]^{\left( {n - 1} \right)} = {1^{\left( {n - 1} \right)}}.\]

We set \(u = y'\), \(v = 1 + {x^2}.\) The left side has the form:

\[ \left[ {y'\left( {1 + {x^2}} \right)} \right]^{\left( {n - 1} \right)} = \sum\limits_{i = 0}^{n - 1} {\left( {\begin{array}{*{20}{c}} {n - 1}\\ i \end{array}} \right){u^{\left( {n - 1 - i} \right)}}{v^{\left( i \right)}}} = \sum\limits_{i = 0}^{n - 1} {\left( {\begin{array}{*{20}{c}} {n - 1}\\ i \end{array}} \right){{\left( {y'} \right)}^{\left( {n - 1 - i} \right)}} \cdot {{\left( {1 + {x^2}} \right)}^{\left( i \right)}}} = {y^{\left( n \right)}}\left( {1 + {x^2}} \right) + \left( {n - 1} \right){y^{\left( {n - 1} \right)}} \cdot 2x + \frac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{2} \cdot {y^{\left( {n - 2} \right)}} \cdot 2 + \ldots \]

The remaining terms in the expansion are obviously zero. Thus, we obtain the following equation:

\[\left( {1 + {x^2}} \right){y^{\left( n \right)}} + 2x\left( {n - 1} \right){y^{\left( {n - 1} \right)}} + \left( {n - 1} \right)\left( {n - 2} \right){y^{\left( {n - 2} \right)}} = 0, \]

which after substitution \(x = 0\) takes the form:

\[{y^{\left( n \right)}}\left( 0 \right) + \left( {n - 1} \right)\left( {n - 2} \right){y^{\left( {n - 2} \right)}}\left( 0 \right) = 0.\]

First we consider the case when the order of the derivative is even. Let \(n = 2k\) \(\left( {k = 0,1,2, \ldots } \right).\) The second derivative of the inverse tangent is given by

\[y^{\prime\prime} = {\left( {\arctan x} \right)^{\prime\prime} = \left( {\frac{1}{{1 + {x^2}}}} \right)'} = - \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} \cdot {\left( {1 + {x^2}} \right)'} = - \frac{2x}{{{{\left( {1 + {x^2}} \right)}^2}}}.\]

when \(x = 0,\) it is equal to zero:

\[y^{\prime\prime}\left( 0 \right) = - \frac{{2 \cdot 0}}{{{{\left( {1 + {0^2}} \right)}^2}}} = 0.\]

Then it follows from the recurrence relationship

\[y^{\left( n \right)}\left( 0 \right) = - \left( {n - 1} \right)\left( {n - 2} \right){y^{\left( {n - 2} \right)}}\left( 0 \right)\]

that all derivatives of the inverse tangent of even order for \(x = 0\) are equal to zero.

For odd \(n\) \(\left( {n = 2k,\;k = 0,1,2, \ldots } \right)\), the recurrence formula can be written in the form:

\[{y^{\left( {2k + 1} \right)}}\left( 0 \right) = - \left( {2k + 1 - 1} \right)\cdot \left( {2k + 1 - 2} \right)\cdot {y^{\left( {2k + 1 - 2} \right)}}\left( 0 \right),\;\; \Rightarrow {y^{\left( {2k + 1} \right)}}\left( 0 \right) = - 2k\left( {2k - 1} \right)\cdot {y^{\left( {2k - 1} \right)}}\left( 0 \right).\]

Given that

\[y'\left( 0 \right) = \frac{1}{{1 + {0^2}}} = 1,\]

we calculate the first few odd derivatives:

\[y^{\prime\prime\prime}\left( 0 \right) = - 2 \cdot 1 \cdot y'\left( 0 \right) = {\left( { - 1} \right)^1} \cdot 1 \cdot 2 = {\left( { - 1} \right)^1} \cdot 2!\]
\[{y^{\left( 5 \right)}}\left( 0 \right) = - 4 \cdot 3 \cdot y^{\prime\prime\prime}\left( 0 \right) = - 4 \cdot 3 \cdot {\left( { - 1} \right)^1} \cdot 2! = {\left( { - 1} \right)^2} \cdot 2! \cdot 3 \cdot 4 = {\left( { - 1} \right)^2} \cdot 4!\]
\[{y^{\left( 7 \right)}}\left( 0 \right) = - 6 \cdot 5 \cdot {y^{\left( 5 \right)}}\left( 0 \right) = - 6 \cdot 5 \cdot {\left( { - 1} \right)^2} \cdot 4! = {\left( { - 1} \right)^3} \cdot 6!\]

Hence, it is easy to establish the general expression for the derivative of the inverse tangent of odd order for \(x = 0:\)

\[y^{\left( {2k + 1} \right)}\left( 0 \right) = {\left( { - 1} \right)^k}\left( {2k} \right)!\]

Example 17.

Calculate the \(4\)th derivative of the function \[y = \frac{{{x^2}}}{{x - 1}}\] at \(x = 2.\)

Solution.

We apply the Leibniz formula

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

Suppose that \(u = \frac{1}{{x - 1}}\) and \(v = {x^2}.\) This yields:

\[y^{\left( 4 \right)} = \left( {\frac{{{x^2}}}{{x - 1}}} \right)^{\left( 4 \right)} = \sum\limits_{i = 0}^4 {\left( {\begin{array}{*{20}{c}} 4\\ i \end{array}} \right){{\left( {\frac{1}{{x - 1}}} \right)}^{\left( {4 - i} \right)}}{{\left( {{x^2}} \right)}^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}} 4\\ 0 \end{array}} \right){\left( {\frac{1}{{x - 1}}} \right)^{\left( 4 \right)}}{x^2} + \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right)\left( {\frac{1}{{x - 1}}} \right)^{\prime\prime\prime}\left( {{x^2}} \right)^\prime + \left( {\begin{array}{*{20}{c}} 4\\ 2 \end{array}} \right)\left( {\frac{1}{{x - 1}}} \right)^{\prime\prime}\left( {{x^2}} \right)^{\prime\prime} + \ldots \]

The derivatives of \(u = \frac{1}{{x - 1}}\) are

\[u^\prime = \left( {\frac{1}{{x - 1}}} \right)^\prime = - \frac{1}{{{{\left( {x - 1} \right)}^2}}};\]
\[u^{\prime\prime} = \left( {\frac{1}{{x - 1}}} \right)^{\prime\prime} = \left( { - \frac{1}{{{{\left( {x - 1} \right)}^2}}}} \right)^\prime = \frac{2}{{{{\left( {x - 1} \right)}^3}}};\]
\[u^{\prime\prime\prime} = \left( {\frac{1}{{x - 1}}} \right)^{\prime\prime\prime} = \left( {\frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right)^\prime = - \frac{6}{{{{\left( {x - 1} \right)}^4}}};\]
\[{u^{\left( 4 \right)}} = {\left( {\frac{1}{{x - 1}}} \right)^{\left( 4 \right)}} = \left( { - \frac{6}{{{{\left( {x - 1} \right)}^4}}}} \right)^\prime = \frac{{24}}{{{{\left( {x - 1} \right)}^5}}}.\]

Similarly we write the derivatives of \(v = {x^2}:\)

\[v^\prime = \left( {{x^2}} \right)^\prime = 2x;\]
\[v^{\prime\prime} = \left( {{x^2}} \right)^{\prime\prime} = \left( {2x} \right)^\prime = 2;\]
\[v^{\prime\prime\prime} = {v^{\left( 4 \right)}} \equiv 0.\]

Hence

\[y^{\left( 4 \right)} = \left( {\begin{array}{*{20}{c}} 4\\ 0 \end{array}} \right) \cdot \frac{{24}}{{{{\left( {x - 1} \right)}^5}}} \cdot {x^2} + \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right) \cdot \left( { - \frac{6}{{{{\left( {x - 1} \right)}^4}}}} \right) \cdot 2x + \left( {\begin{array}{*{20}{c}} 4\\ 2 \end{array}} \right) \cdot \frac{2}{{{{\left( {x - 1} \right)}^3}}} \cdot 2 = 1 \cdot \frac{{24{x^2}}}{{{{\left( {x - 1} \right)}^5}}} - 4 \cdot \frac{{12x}}{{{{\left( {x - 1} \right)}^4}}} + 6 \cdot \frac{4}{{{{\left( {x - 1} \right)}^3}}} = \frac{{24}}{{{{\left( {x - 1} \right)}^5}}}\left[ {{x^2} - 2x\left( {x - 1} \right) + {{\left( {x - 1} \right)}^2}} \right] = \frac{{24}}{{{{\left( {x - 1} \right)}^5}}}\left[ {\cancel{\color{blue}{{x^2}}} - \cancel{\color{blue}{2{x^2}}} + \cancel{\color{red}{2x}} + \cancel{\color{blue}{{x^2}}} - \cancel{\color{red}{2x}} + 1} \right] = \frac{{24}}{{{{\left( {x - 1} \right)}^5}}}.\]

At \(x = 2,\) the fourth derivative is equal to

\[y^{\left( 4 \right)}\left( 2 \right) = \frac{{24}}{{{{\left( {2 - 1} \right)}^5}}} = 24.\]

Example 18.

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

Solution.

We denote \(u = \frac{1}{{x + 2}},\) \(v = {x^3}.\) Using the Leibniz formula

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

we can write

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

Calculate the derivatives of \(u = \frac{1}{{x + 2}}\) and \(v = {x^3}:\)

\[u^\prime = \left( {\frac{1}{{x + 2}}} \right)^\prime = - \frac{1}{{{{\left( {x + 2} \right)}^2}}};\]
\[u^{\prime\prime} = \left( {\frac{1}{{x + 2}}} \right)^{\prime\prime} = \left( { - \frac{1}{{{{\left( {x + 2} \right)}^2}}}} \right)^\prime = \frac{2}{{{{\left( {x + 2} \right)}^3}}};\]
\[u^{\prime\prime\prime} = \left( {\frac{1}{{x + 2}}} \right)^{\prime\prime\prime} = \left( {\frac{2}{{{{\left( {x + 2} \right)}^3}}}} \right)^\prime = - \frac{6}{{{{\left( {x + 2} \right)}^4}}};\]
\[v^\prime = \left( {{x^3}} \right)^\prime = 3{x^2};\]
\[v^{\prime\prime} = \left( {{x^3}} \right)^{\prime\prime} = \left( {3{x^2}} \right)^\prime = 6x;\]
\[v^{\prime\prime\prime} = \left( {{x^3}} \right)^{\prime\prime\prime} = \left( {6x} \right)^\prime = 6.\]

Substitute these expressions for the derivatives in the series expansion:

\[y^{\prime\prime\prime} = 1 \cdot \left( { - \frac{6}{{{{\left( {x + 2} \right)}^4}}}} \right) \cdot {x^3} + 3 \cdot \frac{2}{{{{\left( {x + 2} \right)}^3}}} \cdot 3{x^2} + 3 \cdot \left( { - \frac{1}{{{{\left( {x + 2} \right)}^2}}}} \right) \cdot 6x + 1 \cdot \frac{1}{{x + 2}} \cdot 6 = - \frac{{6{x^3}}}{{{{\left( {x + 2} \right)}^4}}} + \frac{{18{x^2}}}{{{{\left( {x + 2} \right)}^3}}} - \frac{{18x}}{{{{\left( {x + 2} \right)}^2}}} + \frac{6}{{x + 2}} = \frac{6}{{{{\left( {x + 2} \right)}^4}}}\left[ { - {x^3} + 3{x^2}\left( {x + 2} \right) - 3x{{\left( {x + 2} \right)}^2} + {{\left( {x + 2} \right)}^3}} \right] = \frac{6}{{{{\left( {x + 2} \right)}^4}}}\left[ { - {x^3} + 3{x^2}\left( {x + 2} \right) - 3x\left( {{x^2} + 4x + 4} \right) + \left( {{x^3} + 6{x^2} + 12x + 8} \right)} \right] = \frac{6}{{{{\left( {x + 2} \right)}^4}}}\left[ { - \cancel{\color{blue}{{x^3}}} + \cancel{\color{blue}{3{x^3}}} + \cancel{\color{red}{6{x^2}}} - \cancel{\color{blue}{3{x^3}}} - \cancel{\color{red}{12{x^2}}} - \cancel{\color{darkgreen}{12x}} + \cancel{\color{blue}{{x^3}}} + \cancel{\color{red}{6{x^2}}} + \cancel{\color{darkgreen}{12x}} + \color{magenta}{8}} \right] = \frac{48}{{{{\left( {x + 2} \right)}^4}}}.\]

At the point \(x = -1,\) the \(3\)rd derivative of the function is equal to

\[y^{\prime\prime\prime}\left( { - 1} \right) = \frac{48}{{{{\left( { - 1 + 2} \right)}^4}}} = 48.\]
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