Calculus Examples

$h (v) = \sqrt{5 v} + \ln (v^{4}) e^{6 + 9 v}$

Step 1

By the Sum Rule, the derivative of $\sqrt{5 v} + \ln (v^{4}) e^{6 + 9 v}$ with respect to $v$ is $\frac{d}{d v} [\sqrt{5 v}] + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$ .

$\frac{d}{d v} [\sqrt{5 v}] + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2

Evaluate $\frac{d}{d v} [\sqrt{5 v}]$ .

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 v}$ as ${(5 v)}^{\frac{1}{2}}$ .

$\frac{d}{d v} [{(5 v)}^{\frac{1}{2}}] + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.2

Factor $5$ out of $5 v$ .

$\frac{d}{d v} [{(5 (v))}^{\frac{1}{2}}] + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.3

Apply the product rule to $5 (v)$ .

$\frac{d}{d v} [5^{\frac{1}{2}} v^{\frac{1}{2}}] + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.4

Since $5^{\frac{1}{2}}$ is constant with respect to $v$ , the derivative of $5^{\frac{1}{2}} v^{\frac{1}{2}}$ with respect to $v$ is $5^{\frac{1}{2}} \frac{d}{d v} [v^{\frac{1}{2}}]$ .

$5^{\frac{1}{2}} \frac{d}{d v} [v^{\frac{1}{2}}] + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.5

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

$5^{\frac{1}{2}} (\frac{1}{2} v^{\frac{1}{2} - 1}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$5^{\frac{1}{2}} (\frac{1}{2} v^{\frac{1}{2} - 1 \cdot \frac{2}{2}}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.7

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

$5^{\frac{1}{2}} (\frac{1}{2} v^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.8

Combine the numerators over the common denominator.

$5^{\frac{1}{2}} (\frac{1}{2} v^{\frac{1 - 1 \cdot 2}{2}}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.9

Simplify the numerator.

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Step 2.9.1

Multiply $- 1$ by $2$ .

$5^{\frac{1}{2}} (\frac{1}{2} v^{\frac{1 - 2}{2}}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.9.2

Subtract $2$ from $1$ .

$5^{\frac{1}{2}} (\frac{1}{2} v^{\frac{- 1}{2}}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.10

Move the negative in front of the fraction.

$5^{\frac{1}{2}} (\frac{1}{2} v^{- \frac{1}{2}}) + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.11

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

$5^{\frac{1}{2}} \frac{v^{- \frac{1}{2}}}{2} + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.12

Combine $5^{\frac{1}{2}}$ and $\frac{v^{- \frac{1}{2}}}{2}$ .

$\frac{5^{\frac{1}{2}} v^{- \frac{1}{2}}}{2} + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 2.13

Move $v^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$

Step 3

Evaluate $\frac{d}{d v} [\ln (v^{4}) e^{6 + 9 v}]$ .

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Step 3.1

Differentiate using the Product Rule which states that $\frac{d}{d v} [f (v) g (v)]$ is $f (v) \frac{d}{d v} [g (v)] + g (v) \frac{d}{d v} [f (v)]$ where $f (v) = \ln (v^{4})$ and $g (v) = e^{6 + 9 v}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) \frac{d}{d v} [e^{6 + 9 v}] + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d v} [f (g (v))]$ is $f' (g (v)) g' (v)$ where $f (v) = e^{v}$ and $g (v) = 6 + 9 v$ .

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Step 3.2.1

To apply the Chain Rule, set $u_{1}$ as $6 + 9 v$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d v} [6 + 9 v]) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{u_{1}} \frac{d}{d v} [6 + 9 v]) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.2.3

Replace all occurrences of $u_{1}$ with $6 + 9 v$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} \frac{d}{d v} [6 + 9 v]) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.3

By the Sum Rule, the derivative of $6 + 9 v$ with respect to $v$ is $\frac{d}{d v} [6] + \frac{d}{d v} [9 v]$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (\frac{d}{d v} [6] + \frac{d}{d v} [9 v])) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.4

Since $6$ is constant with respect to $v$ , the derivative of $6$ with respect to $v$ is $0$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + \frac{d}{d v} [9 v])) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.5

Since $9$ is constant with respect to $v$ , the derivative of $9 v$ with respect to $v$ is $9 \frac{d}{d v} [v]$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9 \frac{d}{d v} [v])) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.6

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

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9 \cdot 1)) + e^{6 + 9 v} \frac{d}{d v} [\ln (v^{4})]$

Step 3.7

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

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Step 3.7.1

To apply the Chain Rule, set $u_{2}$ as $v^{4}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9 \cdot 1)) + e^{6 + 9 v} (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d v} [v^{4}])$

Step 3.7.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9 \cdot 1)) + e^{6 + 9 v} (\frac{1}{u_{2}} \frac{d}{d v} [v^{4}])$

Step 3.7.3

Replace all occurrences of $u_{2}$ with $v^{4}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9 \cdot 1)) + e^{6 + 9 v} (\frac{1}{v^{4}} \frac{d}{d v} [v^{4}])$

Step 3.8

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

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9 \cdot 1)) + e^{6 + 9 v} (\frac{1}{v^{4}} (4 v^{3}))$

Step 3.9

Multiply $9$ by $1$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} (0 + 9)) + e^{6 + 9 v} (\frac{1}{v^{4}} (4 v^{3}))$

Step 3.10

Add $0$ and $9$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (e^{6 + 9 v} \cdot 9) + e^{6 + 9 v} (\frac{1}{v^{4}} (4 v^{3}))$

Step 3.11

Move $9$ to the left of $e^{6 + 9 v}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 \cdot e^{6 + 9 v}) + e^{6 + 9 v} (\frac{1}{v^{4}} (4 v^{3}))$

Step 3.12

Combine $4$ and $\frac{1}{v^{4}}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + e^{6 + 9 v} (\frac{4}{v^{4}} v^{3})$

Step 3.13

Combine $\frac{4}{v^{4}}$ and $v^{3}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + e^{6 + 9 v} \frac{4 v^{3}}{v^{4}}$

Step 3.14

Cancel the common factor of $v^{3}$ and $v^{4}$ .

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Step 3.14.1

Factor $v^{3}$ out of $4 v^{3}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + e^{6 + 9 v} \frac{v^{3} \cdot 4}{v^{4}}$

Step 3.14.2

Cancel the common factors.

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Step 3.14.2.1

Factor $v^{3}$ out of $v^{4}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + e^{6 + 9 v} \frac{v^{3} \cdot 4}{v^{3} v}$

Step 3.14.2.2

Cancel the common factor.

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + e^{6 + 9 v} \frac{v^{3} \cdot 4}{v^{3} v}$

Step 3.14.2.3

Rewrite the expression.

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + e^{6 + 9 v} \frac{4}{v}$

Step 3.15

Combine $e^{6 + 9 v}$ and $\frac{4}{v}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + \frac{e^{6 + 9 v} \cdot 4}{v}$

Step 3.16

Move $4$ to the left of $e^{6 + 9 v}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \ln (v^{4}) (9 e^{6 + 9 v}) + \frac{4 \cdot e^{6 + 9 v}}{v}$

Step 3.17

To write $\ln (v^{4}) (9 e^{6 + 9 v})$ as a fraction with a common denominator, multiply by $\frac{v}{v}$ .

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \frac{\ln (v^{4}) (9 e^{6 + 9 v}) v}{v} + \frac{4 e^{6 + 9 v}}{v}$

Step 3.18

Combine the numerators over the common denominator.

$\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \frac{\ln (v^{4}) (9 e^{6 + 9 v}) v + 4 e^{6 + 9 v}}{v}$

Step 4

Simplify.

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Step 4.1

Reorder terms.

$\frac{\ln (v^{4}) (9 e^{6 + 9 v}) v + 4 e^{6 + 9 v}}{v} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

Factor $e^{6 + 9 v}$ out of $\ln (v^{4}) (9 e^{6 + 9 v}) v + 4 e^{6 + 9 v}$ .

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Step 4.2.1.1

Factor $e^{6 + 9 v}$ out of $\ln (v^{4}) (9 e^{6 + 9 v}) v$ .

$\frac{e^{6 + 9 v} (\ln (v^{4}) \cdot (9) v) + 4 e^{6 + 9 v}}{v} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$

Step 4.2.1.2

Factor $e^{6 + 9 v}$ out of $4 e^{6 + 9 v}$ .

$\frac{e^{6 + 9 v} (\ln (v^{4}) \cdot (9) v) + e^{6 + 9 v} \cdot 4}{v} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$

Step 4.2.1.3

Factor $e^{6 + 9 v}$ out of $e^{6 + 9 v} (\ln (v^{4}) \cdot (9) v) + e^{6 + 9 v} \cdot 4$ .

$\frac{e^{6 + 9 v} (\ln (v^{4}) \cdot (9) v + 4)}{v} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$

Step 4.2.2

Move $9$ to the left of $\ln (v^{4})$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4)}{v} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$

Step 4.3

To write $\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4)}{v}$ as a fraction with a common denominator, multiply by $\frac{2 v^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4)}{v} \cdot \frac{2 v^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$

Step 4.4

To write $\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$ as a fraction with a common denominator, multiply by $\frac{v}{v}$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4)}{v} \cdot \frac{2 v^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5

Write each expression with a common denominator of $v \cdot 2 v^{\frac{1}{2}}$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.5.1

Multiply $\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4)}{v}$ by $\frac{2 v^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{v (2 v^{\frac{1}{2}})} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5.2

Raise $v$ to the power of $1$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 (v^{1} v^{\frac{1}{2}})} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5.3

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

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{1 + \frac{1}{2}}} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5.4

Write $1$ as a fraction with a common denominator.

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{2}{2} + \frac{1}{2}}} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5.5

Combine the numerators over the common denominator.

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{2 + 1}{2}}} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5.6

Add $2$ and $1$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}} \cdot \frac{v}{v}$

Step 4.5.7

Multiply $\frac{5^{\frac{1}{2}}}{2 v^{\frac{1}{2}}}$ by $\frac{v}{v}$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}} v}{2 v^{\frac{1}{2}} v}$

Step 4.5.8

Raise $v$ to the power of $1$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}} v}{2 (v^{1} v^{\frac{1}{2}})}$

Step 4.5.9

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

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}} v}{2 v^{1 + \frac{1}{2}}}$

Step 4.5.10

Write $1$ as a fraction with a common denominator.

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}} v}{2 v^{\frac{2}{2} + \frac{1}{2}}}$

Step 4.5.11

Combine the numerators over the common denominator.

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}} v}{2 v^{\frac{2 + 1}{2}}}$

Step 4.5.12

Add $2$ and $1$ .

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}} + \frac{5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.6

Combine the numerators over the common denominator.

$\frac{e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}}) + 5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.7

Simplify the numerator.

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Step 4.7.1

Factor $v^{\frac{1}{2}}$ out of $e^{6 + 9 v} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}}) + 5^{\frac{1}{2}} v$ .

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Step 4.7.1.1

Reorder the expression.

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Step 4.7.1.1.1

Reorder $6$ and $9 v$ .

$\frac{e^{9 v + 6} (9 \ln (v^{4}) v + 4) (2 v^{\frac{1}{2}}) + 5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.7.1.1.2

Move $\ln (v^{4})$ .

$\frac{e^{9 v + 6} (9 v \ln (v^{4}) + 4) (2 v^{\frac{1}{2}}) + 5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.7.1.1.3

Move $9 v \ln (v^{4}) + 4$ .

$\frac{e^{9 v + 6} \cdot 2 v^{\frac{1}{2}} (9 v \ln (v^{4}) + 4) + 5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.7.1.1.4

Move $e^{9 v + 6}$ .

$\frac{2 v^{\frac{1}{2}} e^{9 v + 6} (9 v \ln (v^{4}) + 4) + 5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.7.1.2

Factor $v^{\frac{1}{2}}$ out of $2 v^{\frac{1}{2}} e^{9 v + 6} (9 v \ln (v^{4}) + 4)$ .

$\frac{v^{\frac{1}{2}} (2 e^{9 v + 6} (9 v \ln (v^{4}) + 4)) + 5^{\frac{1}{2}} v}{2 v^{\frac{3}{2}}}$

Step 4.7.1.3

Factor $v^{\frac{1}{2}}$ out of $5^{\frac{1}{2}} v$ .

$\frac{v^{\frac{1}{2}} (2 e^{9 v + 6} (9 v \ln (v^{4}) + 4)) + v^{\frac{1}{2}} (5^{\frac{1}{2}} v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}}$

Step 4.7.1.4

Factor $v^{\frac{1}{2}}$ out of $v^{\frac{1}{2}} (2 e^{9 v + 6} (9 v \ln (v^{4}) + 4)) + v^{\frac{1}{2}} (5^{\frac{1}{2}} v^{\frac{1}{2}})$ .

$\frac{v^{\frac{1}{2}} (2 e^{9 v + 6} (9 v \ln (v^{4}) + 4) + 5^{\frac{1}{2}} v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}}$

Step 4.7.2

Apply the distributive property.

$\frac{v^{\frac{1}{2}} (2 e^{9 v + 6} (9 v \ln (v^{4})) + 2 e^{9 v + 6} \cdot 4 + 5^{\frac{1}{2}} v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}}$

Step 4.7.3

Multiply $9$ by $2$ .

$\frac{v^{\frac{1}{2}} (18 e^{9 v + 6} (v \ln (v^{4})) + 2 e^{9 v + 6} \cdot 4 + 5^{\frac{1}{2}} v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}}$

Step 4.7.4

Multiply $4$ by $2$ .

$\frac{v^{\frac{1}{2}} (18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}})}{2 v^{\frac{3}{2}}}$

Step 4.8

Move $v^{\frac{1}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v^{\frac{3}{2}} v^{- \frac{1}{2}}}$

Step 4.9

Simplify the denominator.

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Step 4.9.1

Multiply $v^{\frac{3}{2}}$ by $v^{- \frac{1}{2}}$ by adding the exponents.

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Step 4.9.1.1

Move $v^{- \frac{1}{2}}$ .

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 (v^{- \frac{1}{2}} v^{\frac{3}{2}})}$

Step 4.9.1.2

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

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v^{- \frac{1}{2} + \frac{3}{2}}}$

Step 4.9.1.3

Combine the numerators over the common denominator.

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v^{\frac{- 1 + 3}{2}}}$

Step 4.9.1.4

Add $- 1$ and $3$ .

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v^{\frac{2}{2}}}$

Step 4.9.1.5

Divide $2$ by $2$ .

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v^{1}}$

Step 4.9.2

Simplify $2 v^{1}$ .

$\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v}$

Step 4.10

Reorder factors in $\frac{18 e^{9 v + 6} (v \ln (v^{4})) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v}$ .

$\frac{18 v e^{9 v + 6} \ln (v^{4}) + 8 e^{9 v + 6} + 5^{\frac{1}{2}} v^{\frac{1}{2}}}{2 v}$