Calculus Examples

$t (n) = 3 n^{- \frac{1}{3}} + 4 n^{\frac{5}{3}}$

Step 1

Find the first derivative.

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Step 1.1

By the Sum Rule, the derivative of $3 n^{- \frac{1}{3}} + 4 n^{\frac{5}{3}}$ with respect to $n$ is $\frac{d}{d n} [3 n^{- \frac{1}{3}}] + \frac{d}{d n} [4 n^{\frac{5}{3}}]$ .

$\frac{d}{d n} [3 n^{- \frac{1}{3}}] + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2

Evaluate $\frac{d}{d n} [3 n^{- \frac{1}{3}}]$ .

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Step 1.2.1

Since $3$ is constant with respect to $n$ , the derivative of $3 n^{- \frac{1}{3}}$ with respect to $n$ is $3 \frac{d}{d n} [n^{- \frac{1}{3}}]$ .

$3 \frac{d}{d n} [n^{- \frac{1}{3}}] + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = - \frac{1}{3}$ .

$3 (- \frac{1}{3} n^{- \frac{1}{3} - 1}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$3 (- \frac{1}{3} n^{- \frac{1}{3} - 1 \cdot \frac{3}{3}}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.4

Combine $- 1$ and $\frac{3}{3}$ .

$3 (- \frac{1}{3} n^{- \frac{1}{3} + \frac{- 1 \cdot 3}{3}}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.5

Combine the numerators over the common denominator.

$3 (- \frac{1}{3} n^{\frac{- 1 - 1 \cdot 3}{3}}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.6

Simplify the numerator.

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Step 1.2.6.1

Multiply $- 1$ by $3$ .

$3 (- \frac{1}{3} n^{\frac{- 1 - 3}{3}}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.6.2

Subtract $3$ from $- 1$ .

$3 (- \frac{1}{3} n^{\frac{- 4}{3}}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.7

Move the negative in front of the fraction.

$3 (- \frac{1}{3} n^{- \frac{4}{3}}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.8

Combine $n^{- \frac{4}{3}}$ and $\frac{1}{3}$ .

$3 (- \frac{n^{- \frac{4}{3}}}{3}) + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.9

Multiply $- 1$ by $3$ .

$- 3 \frac{n^{- \frac{4}{3}}}{3} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.10

Combine $- 3$ and $\frac{n^{- \frac{4}{3}}}{3}$ .

$\frac{- 3 n^{- \frac{4}{3}}}{3} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.11

Move $n^{- \frac{4}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{- 3}{3 n^{\frac{4}{3}}} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.12

Factor $3$ out of $- 3$ .

$\frac{3 \cdot - 1}{3 n^{\frac{4}{3}}} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.13

Cancel the common factors.

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Step 1.2.13.1

Factor $3$ out of $3 n^{\frac{4}{3}}$ .

$\frac{3 \cdot - 1}{3 (n^{\frac{4}{3}})} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.13.2

Cancel the common factor.

$\frac{3 \cdot - 1}{3 n^{\frac{4}{3}}} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.13.3

Rewrite the expression.

$\frac{- 1}{n^{\frac{4}{3}}} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.2.14

Move the negative in front of the fraction.

$- \frac{1}{n^{\frac{4}{3}}} + \frac{d}{d n} [4 n^{\frac{5}{3}}]$

Step 1.3

Evaluate $\frac{d}{d n} [4 n^{\frac{5}{3}}]$ .

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Step 1.3.1

Since $4$ is constant with respect to $n$ , the derivative of $4 n^{\frac{5}{3}}$ with respect to $n$ is $4 \frac{d}{d n} [n^{\frac{5}{3}}]$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 \frac{d}{d n} [n^{\frac{5}{3}}]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{5}{3}$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 (\frac{5}{3} n^{\frac{5}{3} - 1})$

Step 1.3.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 (\frac{5}{3} n^{\frac{5}{3} - 1 \cdot \frac{3}{3}})$

Step 1.3.4

Combine $- 1$ and $\frac{3}{3}$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 (\frac{5}{3} n^{\frac{5}{3} + \frac{- 1 \cdot 3}{3}})$

Step 1.3.5

Combine the numerators over the common denominator.

$- \frac{1}{n^{\frac{4}{3}}} + 4 (\frac{5}{3} n^{\frac{5 - 1 \cdot 3}{3}})$

Step 1.3.6

Simplify the numerator.

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Step 1.3.6.1

Multiply $- 1$ by $3$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 (\frac{5}{3} n^{\frac{5 - 3}{3}})$

Step 1.3.6.2

Subtract $3$ from $5$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 (\frac{5}{3} n^{\frac{2}{3}})$

Step 1.3.7

Combine $\frac{5}{3}$ and $n^{\frac{2}{3}}$ .

$- \frac{1}{n^{\frac{4}{3}}} + 4 \frac{5 n^{\frac{2}{3}}}{3}$

Step 1.3.8

Combine $4$ and $\frac{5 n^{\frac{2}{3}}}{3}$ .

$- \frac{1}{n^{\frac{4}{3}}} + \frac{4 (5 n^{\frac{2}{3}})}{3}$

Step 1.3.9

Multiply $5$ by $4$ .

$- \frac{1}{n^{\frac{4}{3}}} + \frac{20 n^{\frac{2}{3}}}{3}$

Step 1.4

Reorder terms.

$f' (n) = \frac{20 n^{\frac{2}{3}}}{3} - \frac{1}{n^{\frac{4}{3}}}$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $\frac{20 n^{\frac{2}{3}}}{3} - \frac{1}{n^{\frac{4}{3}}}$ with respect to $n$ is $\frac{d}{d n} [\frac{20 n^{\frac{2}{3}}}{3}] + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$ .

$\frac{d}{d n} [\frac{20 n^{\frac{2}{3}}}{3}] + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2

Evaluate $\frac{d}{d n} [\frac{20 n^{\frac{2}{3}}}{3}]$ .

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Step 2.2.1

Since $\frac{20}{3}$ is constant with respect to $n$ , the derivative of $\frac{20 n^{\frac{2}{3}}}{3}$ with respect to $n$ is $\frac{20}{3} \frac{d}{d n} [n^{\frac{2}{3}}]$ .

$\frac{20}{3} \frac{d}{d n} [n^{\frac{2}{3}}] + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{2}{3}$ .

$\frac{20}{3} (\frac{2}{3} n^{\frac{2}{3} - 1}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{20}{3} (\frac{2}{3} n^{\frac{2}{3} - 1 \cdot \frac{3}{3}}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.4

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{20}{3} (\frac{2}{3} n^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.5

Combine the numerators over the common denominator.

$\frac{20}{3} (\frac{2}{3} n^{\frac{2 - 1 \cdot 3}{3}}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.6

Simplify the numerator.

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Step 2.2.6.1

Multiply $- 1$ by $3$ .

$\frac{20}{3} (\frac{2}{3} n^{\frac{2 - 3}{3}}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.6.2

Subtract $3$ from $2$ .

$\frac{20}{3} (\frac{2}{3} n^{\frac{- 1}{3}}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.7

Move the negative in front of the fraction.

$\frac{20}{3} (\frac{2}{3} n^{- \frac{1}{3}}) + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.8

Combine $\frac{2}{3}$ and $n^{- \frac{1}{3}}$ .

$\frac{20}{3} \cdot \frac{2 n^{- \frac{1}{3}}}{3} + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.9

Multiply $\frac{20}{3}$ by $\frac{2 n^{- \frac{1}{3}}}{3}$ .

$\frac{20 (2 n^{- \frac{1}{3}})}{3 \cdot 3} + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.10

Multiply $2$ by $20$ .

$\frac{40 n^{- \frac{1}{3}}}{3 \cdot 3} + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.11

Multiply $3$ by $3$ .

$\frac{40 n^{- \frac{1}{3}}}{9} + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.2.12

Move $n^{- \frac{1}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} + \frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$

Step 2.3

Evaluate $\frac{d}{d n} [- \frac{1}{n^{\frac{4}{3}}}]$ .

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Step 2.3.1

Differentiate using the Product Rule which states that $\frac{d}{d n} [f (n) g (n)]$ is $f (n) \frac{d}{d n} [g (n)] + g (n) \frac{d}{d n} [f (n)]$ where $f (n) = - 1$ and $g (n) = \frac{1}{n^{\frac{4}{3}}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - \frac{d}{d n} [\frac{1}{n^{\frac{4}{3}}}] + \frac{1}{n^{\frac{4}{3}}} \frac{d}{d n} [- 1]$

Step 2.3.2

Rewrite $\frac{1}{n^{\frac{4}{3}}}$ as ${(n^{\frac{4}{3}})}^{- 1}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - \frac{d}{d n} [{(n^{\frac{4}{3}})}^{- 1}] + \frac{1}{n^{\frac{4}{3}}} \frac{d}{d n} [- 1]$

Step 2.3.3

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = n^{- 1}$ and $g (n) = n^{\frac{4}{3}}$ .

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Step 2.3.3.1

To apply the Chain Rule, set $u$ as $n^{\frac{4}{3}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (\frac{d}{d u} [u^{- 1}] \frac{d}{d n} [n^{\frac{4}{3}}]) + \frac{1}{n^{\frac{4}{3}}} \frac{d}{d n} [- 1]$

Step 2.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 1$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- u^{- 2} \frac{d}{d n} [n^{\frac{4}{3}}]) + \frac{1}{n^{\frac{4}{3}}} \frac{d}{d n} [- 1]$

Step 2.3.3.3

Replace all occurrences of $u$ with $n^{\frac{4}{3}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- {(n^{\frac{4}{3}})}^{- 2} \frac{d}{d n} [n^{\frac{4}{3}}]) + \frac{1}{n^{\frac{4}{3}}} \frac{d}{d n} [- 1]$

Step 2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{4}{3}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- {(n^{\frac{4}{3}})}^{- 2} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{1}{n^{\frac{4}{3}}} \frac{d}{d n} [- 1]$

Step 2.3.5

Since $- 1$ is constant with respect to $n$ , the derivative of $- 1$ with respect to $n$ is $0$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- {(n^{\frac{4}{3}})}^{- 2} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.6

Multiply the exponents in ${(n^{\frac{4}{3}})}^{- 2}$ .

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Step 2.3.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{\frac{4}{3} \cdot - 2} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.6.2

Multiply $\frac{4}{3} \cdot - 2$ .

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Step 2.3.6.2.1

Combine $\frac{4}{3}$ and $- 2$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{\frac{4 \cdot - 2}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.6.2.2

Multiply $4$ by $- 2$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{\frac{- 8}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.6.3

Move the negative in front of the fraction.

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1 \cdot \frac{3}{3}})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.8

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4}{3} + \frac{- 1 \cdot 3}{3}})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.9

Combine the numerators over the common denominator.

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4 - 1 \cdot 3}{3}})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.10

Simplify the numerator.

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Step 2.3.10.1

Multiply $- 1$ by $3$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4 - 3}{3}})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.10.2

Subtract $3$ from $4$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{1}{3}})) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.11

Combine $\frac{4}{3}$ and $n^{\frac{1}{3}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - (- n^{- \frac{8}{3}} \frac{4 n^{\frac{1}{3}}}{3}) + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.12

Combine $\frac{4 n^{\frac{1}{3}}}{3}$ and $n^{- \frac{8}{3}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4 n^{\frac{1}{3}} n^{- \frac{8}{3}}}{3} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.13

Multiply $n^{\frac{1}{3}}$ by $n^{- \frac{8}{3}}$ by adding the exponents.

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Step 2.3.13.1

Move $n^{- \frac{8}{3}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4 (n^{- \frac{8}{3}} n^{\frac{1}{3}})}{3} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.13.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4 n^{- \frac{8}{3} + \frac{1}{3}}}{3} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.13.3

Combine the numerators over the common denominator.

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4 n^{\frac{- 8 + 1}{3}}}{3} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.13.4

Add $- 8$ and $1$ .

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4 n^{\frac{- 7}{3}}}{3} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.13.5

Move the negative in front of the fraction.

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4 n^{- \frac{7}{3}}}{3} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.14

Move $n^{- \frac{7}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{40}{9 n^{\frac{1}{3}}} - - \frac{4}{3 n^{\frac{7}{3}}} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.15

Multiply $- 1$ by $- 1$ .

$\frac{40}{9 n^{\frac{1}{3}}} + 1 \frac{4}{3 n^{\frac{7}{3}}} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.16

Multiply $\frac{4}{3 n^{\frac{7}{3}}}$ by $1$ .

$\frac{40}{9 n^{\frac{1}{3}}} + \frac{4}{3 n^{\frac{7}{3}}} + \frac{1}{n^{\frac{4}{3}}} \cdot 0$

Step 2.3.17

Multiply $\frac{1}{n^{\frac{4}{3}}}$ by $0$ .

$\frac{40}{9 n^{\frac{1}{3}}} + \frac{4}{3 n^{\frac{7}{3}}} + 0$

Step 2.3.18

Add $\frac{4}{3 n^{\frac{7}{3}}}$ and $0$ .

$f'' (n) = \frac{40}{9 n^{\frac{1}{3}}} + \frac{4}{3 n^{\frac{7}{3}}}$

Step 3

Find the third derivative.

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Step 3.1

By the Sum Rule, the derivative of $\frac{40}{9 n^{\frac{1}{3}}} + \frac{4}{3 n^{\frac{7}{3}}}$ with respect to $n$ is $\frac{d}{d n} [\frac{40}{9 n^{\frac{1}{3}}}] + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$ .

$\frac{d}{d n} [\frac{40}{9 n^{\frac{1}{3}}}] + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2

Evaluate $\frac{d}{d n} [\frac{40}{9 n^{\frac{1}{3}}}]$ .

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Step 3.2.1

Since $\frac{40}{9}$ is constant with respect to $n$ , the derivative of $\frac{40}{9 n^{\frac{1}{3}}}$ with respect to $n$ is $\frac{40}{9} \frac{d}{d n} [\frac{1}{n^{\frac{1}{3}}}]$ .

$\frac{40}{9} \frac{d}{d n} [\frac{1}{n^{\frac{1}{3}}}] + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.2

Rewrite $\frac{1}{n^{\frac{1}{3}}}$ as ${(n^{\frac{1}{3}})}^{- 1}$ .

$\frac{40}{9} \frac{d}{d n} [{(n^{\frac{1}{3}})}^{- 1}] + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.3

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = n^{- 1}$ and $g (n) = n^{\frac{1}{3}}$ .

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Step 3.2.3.1

To apply the Chain Rule, set $u_{1}$ as $n^{\frac{1}{3}}$ .

$\frac{40}{9} (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d n} [n^{\frac{1}{3}}]) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = - 1$ .

$\frac{40}{9} (- {u_{1}}^{- 2} \frac{d}{d n} [n^{\frac{1}{3}}]) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.3.3

Replace all occurrences of $u_{1}$ with $n^{\frac{1}{3}}$ .

$\frac{40}{9} (- {(n^{\frac{1}{3}})}^{- 2} \frac{d}{d n} [n^{\frac{1}{3}}]) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.4

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{1}{3}$ .

$\frac{40}{9} (- {(n^{\frac{1}{3}})}^{- 2} (\frac{1}{3} n^{\frac{1}{3} - 1})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.5

Multiply the exponents in ${(n^{\frac{1}{3}})}^{- 2}$ .

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Step 3.2.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{40}{9} (- n^{\frac{1}{3} \cdot - 2} (\frac{1}{3} n^{\frac{1}{3} - 1})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.5.2

Combine $\frac{1}{3}$ and $- 2$ .

$\frac{40}{9} (- n^{\frac{- 2}{3}} (\frac{1}{3} n^{\frac{1}{3} - 1})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.5.3

Move the negative in front of the fraction.

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{\frac{1}{3} - 1})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{\frac{1}{3} - 1 \cdot \frac{3}{3}})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.7

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.8

Combine the numerators over the common denominator.

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{\frac{1 - 1 \cdot 3}{3}})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.9

Simplify the numerator.

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Step 3.2.9.1

Multiply $- 1$ by $3$ .

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{\frac{1 - 3}{3}})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.9.2

Subtract $3$ from $1$ .

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{\frac{- 2}{3}})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.10

Move the negative in front of the fraction.

$\frac{40}{9} (- n^{- \frac{2}{3}} (\frac{1}{3} n^{- \frac{2}{3}})) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.11

Combine $\frac{1}{3}$ and $n^{- \frac{2}{3}}$ .

$\frac{40}{9} (- n^{- \frac{2}{3}} \frac{n^{- \frac{2}{3}}}{3}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.12

Combine $\frac{n^{- \frac{2}{3}}}{3}$ and $n^{- \frac{2}{3}}$ .

$\frac{40}{9} (- \frac{n^{- \frac{2}{3}} n^{- \frac{2}{3}}}{3}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.13

Multiply $n^{- \frac{2}{3}}$ by $n^{- \frac{2}{3}}$ by adding the exponents.

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Step 3.2.13.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{40}{9} (- \frac{n^{- \frac{2}{3} - \frac{2}{3}}}{3}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.13.2

Combine the numerators over the common denominator.

$\frac{40}{9} (- \frac{n^{\frac{- 2 - 2}{3}}}{3}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.13.3

Subtract $2$ from $- 2$ .

$\frac{40}{9} (- \frac{n^{\frac{- 4}{3}}}{3}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.13.4

Move the negative in front of the fraction.

$\frac{40}{9} (- \frac{n^{- \frac{4}{3}}}{3}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.14

Move $n^{- \frac{4}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{40}{9} (- \frac{1}{3 n^{\frac{4}{3}}}) + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.15

Multiply $\frac{40}{9}$ by $\frac{1}{3 n^{\frac{4}{3}}}$ .

$- \frac{40}{9 (3 n^{\frac{4}{3}})} + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.2.16

Multiply $3$ by $9$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$

Step 3.3

Evaluate $\frac{d}{d n} [\frac{4}{3 n^{\frac{7}{3}}}]$ .

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Step 3.3.1

Since $\frac{4}{3}$ is constant with respect to $n$ , the derivative of $\frac{4}{3 n^{\frac{7}{3}}}$ with respect to $n$ is $\frac{4}{3} \frac{d}{d n} [\frac{1}{n^{\frac{7}{3}}}]$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} \frac{d}{d n} [\frac{1}{n^{\frac{7}{3}}}]$

Step 3.3.2

Rewrite $\frac{1}{n^{\frac{7}{3}}}$ as ${(n^{\frac{7}{3}})}^{- 1}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} \frac{d}{d n} [{(n^{\frac{7}{3}})}^{- 1}]$

Step 3.3.3

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = n^{- 1}$ and $g (n) = n^{\frac{7}{3}}$ .

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Step 3.3.3.1

To apply the Chain Rule, set $u_{2}$ as $n^{\frac{7}{3}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d n} [n^{\frac{7}{3}}])$

Step 3.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 1$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- {u_{2}}^{- 2} \frac{d}{d n} [n^{\frac{7}{3}}])$

Step 3.3.3.3

Replace all occurrences of $u_{2}$ with $n^{\frac{7}{3}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- {(n^{\frac{7}{3}})}^{- 2} \frac{d}{d n} [n^{\frac{7}{3}}])$

Step 3.3.4

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{7}{3}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- {(n^{\frac{7}{3}})}^{- 2} (\frac{7}{3} n^{\frac{7}{3} - 1}))$

Step 3.3.5

Multiply the exponents in ${(n^{\frac{7}{3}})}^{- 2}$ .

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Step 3.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{\frac{7}{3} \cdot - 2} (\frac{7}{3} n^{\frac{7}{3} - 1}))$

Step 3.3.5.2

Multiply $\frac{7}{3} \cdot - 2$ .

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Step 3.3.5.2.1

Combine $\frac{7}{3}$ and $- 2$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{\frac{7 \cdot - 2}{3}} (\frac{7}{3} n^{\frac{7}{3} - 1}))$

Step 3.3.5.2.2

Multiply $7$ by $- 2$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{\frac{- 14}{3}} (\frac{7}{3} n^{\frac{7}{3} - 1}))$

Step 3.3.5.3

Move the negative in front of the fraction.

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} (\frac{7}{3} n^{\frac{7}{3} - 1}))$

Step 3.3.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} (\frac{7}{3} n^{\frac{7}{3} - 1 \cdot \frac{3}{3}}))$

Step 3.3.7

Combine $- 1$ and $\frac{3}{3}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} (\frac{7}{3} n^{\frac{7}{3} + \frac{- 1 \cdot 3}{3}}))$

Step 3.3.8

Combine the numerators over the common denominator.

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} (\frac{7}{3} n^{\frac{7 - 1 \cdot 3}{3}}))$

Step 3.3.9

Simplify the numerator.

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Step 3.3.9.1

Multiply $- 1$ by $3$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} (\frac{7}{3} n^{\frac{7 - 3}{3}}))$

Step 3.3.9.2

Subtract $3$ from $7$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} (\frac{7}{3} n^{\frac{4}{3}}))$

Step 3.3.10

Combine $\frac{7}{3}$ and $n^{\frac{4}{3}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- n^{- \frac{14}{3}} \frac{7 n^{\frac{4}{3}}}{3})$

Step 3.3.11

Combine $\frac{7 n^{\frac{4}{3}}}{3}$ and $n^{- \frac{14}{3}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7 n^{\frac{4}{3}} n^{- \frac{14}{3}}}{3})$

Step 3.3.12

Multiply $n^{\frac{4}{3}}$ by $n^{- \frac{14}{3}}$ by adding the exponents.

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Step 3.3.12.1

Move $n^{- \frac{14}{3}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7 (n^{- \frac{14}{3}} n^{\frac{4}{3}})}{3})$

Step 3.3.12.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7 n^{- \frac{14}{3} + \frac{4}{3}}}{3})$

Step 3.3.12.3

Combine the numerators over the common denominator.

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7 n^{\frac{- 14 + 4}{3}}}{3})$

Step 3.3.12.4

Add $- 14$ and $4$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7 n^{\frac{- 10}{3}}}{3})$

Step 3.3.12.5

Move the negative in front of the fraction.

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7 n^{- \frac{10}{3}}}{3})$

Step 3.3.13

Move $n^{- \frac{10}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} + \frac{4}{3} (- \frac{7}{3 n^{\frac{10}{3}}})$

Step 3.3.14

Multiply $\frac{4}{3}$ by $\frac{7}{3 n^{\frac{10}{3}}}$ .

$- \frac{40}{27 n^{\frac{4}{3}}} - \frac{4 \cdot 7}{3 (3 n^{\frac{10}{3}})}$

Step 3.3.15

Multiply $4$ by $7$ .

$- \frac{40}{27 n^{\frac{4}{3}}} - \frac{28}{3 (3 n^{\frac{10}{3}})}$

Step 3.3.16

Multiply $3$ by $3$ .

$f''' (n) = - \frac{40}{27 n^{\frac{4}{3}}} - \frac{28}{9 n^{\frac{10}{3}}}$

Step 4

Find the fourth derivative.

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Step 4.1

By the Sum Rule, the derivative of $- \frac{40}{27 n^{\frac{4}{3}}} - \frac{28}{9 n^{\frac{10}{3}}}$ with respect to $n$ is $\frac{d}{d n} [- \frac{40}{27 n^{\frac{4}{3}}}] + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$ .

$\frac{d}{d n} [- \frac{40}{27 n^{\frac{4}{3}}}] + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2

Evaluate $\frac{d}{d n} [- \frac{40}{27 n^{\frac{4}{3}}}]$ .

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Step 4.2.1

Since $- \frac{40}{27}$ is constant with respect to $n$ , the derivative of $- \frac{40}{27 n^{\frac{4}{3}}}$ with respect to $n$ is $- \frac{40}{27} \frac{d}{d n} [\frac{1}{n^{\frac{4}{3}}}]$ .

$- \frac{40}{27} \frac{d}{d n} [\frac{1}{n^{\frac{4}{3}}}] + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.2

Rewrite $\frac{1}{n^{\frac{4}{3}}}$ as ${(n^{\frac{4}{3}})}^{- 1}$ .

$- \frac{40}{27} \frac{d}{d n} [{(n^{\frac{4}{3}})}^{- 1}] + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.3

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = n^{- 1}$ and $g (n) = n^{\frac{4}{3}}$ .

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Step 4.2.3.1

To apply the Chain Rule, set $u_{1}$ as $n^{\frac{4}{3}}$ .

$- \frac{40}{27} (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d n} [n^{\frac{4}{3}}]) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = - 1$ .

$- \frac{40}{27} (- {u_{1}}^{- 2} \frac{d}{d n} [n^{\frac{4}{3}}]) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.3.3

Replace all occurrences of $u_{1}$ with $n^{\frac{4}{3}}$ .

$- \frac{40}{27} (- {(n^{\frac{4}{3}})}^{- 2} \frac{d}{d n} [n^{\frac{4}{3}}]) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.4

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{4}{3}$ .

$- \frac{40}{27} (- {(n^{\frac{4}{3}})}^{- 2} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.5

Multiply the exponents in ${(n^{\frac{4}{3}})}^{- 2}$ .

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Step 4.2.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{40}{27} (- n^{\frac{4}{3} \cdot - 2} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.5.2

Multiply $\frac{4}{3} \cdot - 2$ .

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Step 4.2.5.2.1

Combine $\frac{4}{3}$ and $- 2$ .

$- \frac{40}{27} (- n^{\frac{4 \cdot - 2}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.5.2.2

Multiply $4$ by $- 2$ .

$- \frac{40}{27} (- n^{\frac{- 8}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.5.3

Move the negative in front of the fraction.

$- \frac{40}{27} (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{40}{27} (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4}{3} - 1 \cdot \frac{3}{3}})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.7

Combine $- 1$ and $\frac{3}{3}$ .

$- \frac{40}{27} (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4}{3} + \frac{- 1 \cdot 3}{3}})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.8

Combine the numerators over the common denominator.

$- \frac{40}{27} (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4 - 1 \cdot 3}{3}})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.9

Simplify the numerator.

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Step 4.2.9.1

Multiply $- 1$ by $3$ .

$- \frac{40}{27} (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{4 - 3}{3}})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.9.2

Subtract $3$ from $4$ .

$- \frac{40}{27} (- n^{- \frac{8}{3}} (\frac{4}{3} n^{\frac{1}{3}})) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.10

Combine $\frac{4}{3}$ and $n^{\frac{1}{3}}$ .

$- \frac{40}{27} (- n^{- \frac{8}{3}} \frac{4 n^{\frac{1}{3}}}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.11

Combine $\frac{4 n^{\frac{1}{3}}}{3}$ and $n^{- \frac{8}{3}}$ .

$- \frac{40}{27} (- \frac{4 n^{\frac{1}{3}} n^{- \frac{8}{3}}}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.12

Multiply $n^{\frac{1}{3}}$ by $n^{- \frac{8}{3}}$ by adding the exponents.

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Step 4.2.12.1

Move $n^{- \frac{8}{3}}$ .

$- \frac{40}{27} (- \frac{4 (n^{- \frac{8}{3}} n^{\frac{1}{3}})}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.12.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{40}{27} (- \frac{4 n^{- \frac{8}{3} + \frac{1}{3}}}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.12.3

Combine the numerators over the common denominator.

$- \frac{40}{27} (- \frac{4 n^{\frac{- 8 + 1}{3}}}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.12.4

Add $- 8$ and $1$ .

$- \frac{40}{27} (- \frac{4 n^{\frac{- 7}{3}}}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.12.5

Move the negative in front of the fraction.

$- \frac{40}{27} (- \frac{4 n^{- \frac{7}{3}}}{3}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.13

Move $n^{- \frac{7}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{40}{27} (- \frac{4}{3 n^{\frac{7}{3}}}) + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.14

Multiply $- 1$ by $- 1$ .

$1 (\frac{40}{27}) \frac{4}{3 n^{\frac{7}{3}}} + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.15

Multiply $\frac{40}{27}$ by $1$ .

$\frac{40}{27} \cdot \frac{4}{3 n^{\frac{7}{3}}} + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.16

Multiply $\frac{40}{27}$ by $\frac{4}{3 n^{\frac{7}{3}}}$ .

$\frac{40 \cdot 4}{27 (3 n^{\frac{7}{3}})} + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.17

Multiply $40$ by $4$ .

$\frac{160}{27 (3 n^{\frac{7}{3}})} + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.2.18

Multiply $3$ by $27$ .

$\frac{160}{81 n^{\frac{7}{3}}} + \frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$

Step 4.3

Evaluate $\frac{d}{d n} [- \frac{28}{9 n^{\frac{10}{3}}}]$ .

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Step 4.3.1

Since $- \frac{28}{9}$ is constant with respect to $n$ , the derivative of $- \frac{28}{9 n^{\frac{10}{3}}}$ with respect to $n$ is $- \frac{28}{9} \frac{d}{d n} [\frac{1}{n^{\frac{10}{3}}}]$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} \frac{d}{d n} [\frac{1}{n^{\frac{10}{3}}}]$

Step 4.3.2

Rewrite $\frac{1}{n^{\frac{10}{3}}}$ as ${(n^{\frac{10}{3}})}^{- 1}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} \frac{d}{d n} [{(n^{\frac{10}{3}})}^{- 1}]$

Step 4.3.3

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = n^{- 1}$ and $g (n) = n^{\frac{10}{3}}$ .

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Step 4.3.3.1

To apply the Chain Rule, set $u_{2}$ as $n^{\frac{10}{3}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d n} [n^{\frac{10}{3}}])$

Step 4.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 1$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- {u_{2}}^{- 2} \frac{d}{d n} [n^{\frac{10}{3}}])$

Step 4.3.3.3

Replace all occurrences of $u_{2}$ with $n^{\frac{10}{3}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- {(n^{\frac{10}{3}})}^{- 2} \frac{d}{d n} [n^{\frac{10}{3}}])$

Step 4.3.4

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = \frac{10}{3}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- {(n^{\frac{10}{3}})}^{- 2} (\frac{10}{3} n^{\frac{10}{3} - 1}))$

Step 4.3.5

Multiply the exponents in ${(n^{\frac{10}{3}})}^{- 2}$ .

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Step 4.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{\frac{10}{3} \cdot - 2} (\frac{10}{3} n^{\frac{10}{3} - 1}))$

Step 4.3.5.2

Multiply $\frac{10}{3} \cdot - 2$ .

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Step 4.3.5.2.1

Combine $\frac{10}{3}$ and $- 2$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{\frac{10 \cdot - 2}{3}} (\frac{10}{3} n^{\frac{10}{3} - 1}))$

Step 4.3.5.2.2

Multiply $10$ by $- 2$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{\frac{- 20}{3}} (\frac{10}{3} n^{\frac{10}{3} - 1}))$

Step 4.3.5.3

Move the negative in front of the fraction.

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} (\frac{10}{3} n^{\frac{10}{3} - 1}))$

Step 4.3.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} (\frac{10}{3} n^{\frac{10}{3} - 1 \cdot \frac{3}{3}}))$

Step 4.3.7

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} (\frac{10}{3} n^{\frac{10}{3} + \frac{- 1 \cdot 3}{3}}))$

Step 4.3.8

Combine the numerators over the common denominator.

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} (\frac{10}{3} n^{\frac{10 - 1 \cdot 3}{3}}))$

Step 4.3.9

Simplify the numerator.

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Step 4.3.9.1

Multiply $- 1$ by $3$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} (\frac{10}{3} n^{\frac{10 - 3}{3}}))$

Step 4.3.9.2

Subtract $3$ from $10$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} (\frac{10}{3} n^{\frac{7}{3}}))$

Step 4.3.10

Combine $\frac{10}{3}$ and $n^{\frac{7}{3}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- n^{- \frac{20}{3}} \frac{10 n^{\frac{7}{3}}}{3})$

Step 4.3.11

Combine $\frac{10 n^{\frac{7}{3}}}{3}$ and $n^{- \frac{20}{3}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10 n^{\frac{7}{3}} n^{- \frac{20}{3}}}{3})$

Step 4.3.12

Multiply $n^{\frac{7}{3}}$ by $n^{- \frac{20}{3}}$ by adding the exponents.

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Step 4.3.12.1

Move $n^{- \frac{20}{3}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10 (n^{- \frac{20}{3}} n^{\frac{7}{3}})}{3})$

Step 4.3.12.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10 n^{- \frac{20}{3} + \frac{7}{3}}}{3})$

Step 4.3.12.3

Combine the numerators over the common denominator.

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10 n^{\frac{- 20 + 7}{3}}}{3})$

Step 4.3.12.4

Add $- 20$ and $7$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10 n^{\frac{- 13}{3}}}{3})$

Step 4.3.12.5

Move the negative in front of the fraction.

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10 n^{- \frac{13}{3}}}{3})$

Step 4.3.13

Move $n^{- \frac{13}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} - \frac{28}{9} (- \frac{10}{3 n^{\frac{13}{3}}})$

Step 4.3.14

Multiply $- 1$ by $- 1$ .

$\frac{160}{81 n^{\frac{7}{3}}} + 1 (\frac{28}{9}) \frac{10}{3 n^{\frac{13}{3}}}$

Step 4.3.15

Multiply $\frac{28}{9}$ by $1$ .

$\frac{160}{81 n^{\frac{7}{3}}} + \frac{28}{9} \cdot \frac{10}{3 n^{\frac{13}{3}}}$

Step 4.3.16

Multiply $\frac{28}{9}$ by $\frac{10}{3 n^{\frac{13}{3}}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} + \frac{28 \cdot 10}{9 (3 n^{\frac{13}{3}})}$

Step 4.3.17

Multiply $28$ by $10$ .

$\frac{160}{81 n^{\frac{7}{3}}} + \frac{280}{9 (3 n^{\frac{13}{3}})}$

Step 4.3.18

Multiply $3$ by $9$ .

$f^{4} (n) = \frac{160}{81 n^{\frac{7}{3}}} + \frac{280}{27 n^{\frac{13}{3}}}$

Step 5

The fourth derivative of $t (n)$ with respect to $n$ is $\frac{160}{81 n^{\frac{7}{3}}} + \frac{280}{27 n^{\frac{13}{3}}}$ .

$\frac{160}{81 n^{\frac{7}{3}}} + \frac{280}{27 n^{\frac{13}{3}}}$