# 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

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