Calculus Examples

$v^{2} \sqrt{2 v - 5}$

Step 1

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

$\frac{d}{d v} [v^{2} {(2 v - 5)}^{\frac{1}{2}}]$

Step 2

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 $f (v) = v^{2}$ and $g (v) = {(2 v - 5)}^{\frac{1}{2}}$ .

$v^{2} \frac{d}{d v} [{(2 v - 5)}^{\frac{1}{2}}] + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d v} [f (g (v))]$ is $f' (g (v)) g' (v)$ where $f (v) = v^{\frac{1}{2}}$ and $g (v) = 2 v - 5$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $2 v - 5$ .

$v^{2} (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 3.2

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

$v^{2} (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 3.3

Replace all occurrences of $u$ with $2 v - 5$ .

$v^{2} (\frac{1}{2} {(2 v - 5)}^{\frac{1}{2} - 1} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 4

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

$v^{2} (\frac{1}{2} {(2 v - 5)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 5

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

$v^{2} (\frac{1}{2} {(2 v - 5)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 6

Combine the numerators over the common denominator.

$v^{2} (\frac{1}{2} {(2 v - 5)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$v^{2} (\frac{1}{2} {(2 v - 5)}^{\frac{1 - 2}{2}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 7.2

Subtract $2$ from $1$ .

$v^{2} (\frac{1}{2} {(2 v - 5)}^{\frac{- 1}{2}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 8

Combine fractions.

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Step 8.1

Move the negative in front of the fraction.

$v^{2} (\frac{1}{2} {(2 v - 5)}^{- \frac{1}{2}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 8.2

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

$v^{2} (\frac{{(2 v - 5)}^{- \frac{1}{2}}}{2} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 8.3

Move ${(2 v - 5)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$v^{2} (\frac{1}{2 {(2 v - 5)}^{\frac{1}{2}}} \frac{d}{d v} [2 v - 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 8.4

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

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} \frac{d}{d v} [2 v - 5] + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 9

By the Sum Rule, the derivative of $2 v - 5$ with respect to $v$ is $\frac{d}{d v} [2 v] + \frac{d}{d v} [- 5]$ .

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} (\frac{d}{d v} [2 v] + \frac{d}{d v} [- 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 10

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

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} (2 \frac{d}{d v} [v] + \frac{d}{d v} [- 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 11

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

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} (2 \cdot 1 + \frac{d}{d v} [- 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 12

Multiply $2$ by $1$ .

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} (2 + \frac{d}{d v} [- 5]) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 13

Since $- 5$ is constant with respect to $v$ , the derivative of $- 5$ with respect to $v$ is $0$ .

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} (2 + 0) + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 14

Simplify terms.

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Step 14.1

Add $2$ and $0$ .

$\frac{v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} \cdot 2 + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 14.2

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

$\frac{v^{2} \cdot 2}{2 {(2 v - 5)}^{\frac{1}{2}}} + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 14.3

Move $2$ to the left of $v^{2}$ .

$\frac{2 \cdot v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 14.4

Cancel the common factor.

$\frac{2 v^{2}}{2 {(2 v - 5)}^{\frac{1}{2}}} + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 14.5

Rewrite the expression.

$\frac{v^{2}}{{(2 v - 5)}^{\frac{1}{2}}} + {(2 v - 5)}^{\frac{1}{2}} \frac{d}{d v} [v^{2}]$

Step 15

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

$\frac{v^{2}}{{(2 v - 5)}^{\frac{1}{2}}} + {(2 v - 5)}^{\frac{1}{2}} (2 v)$

Step 16

Move $2$ to the left of ${(2 v - 5)}^{\frac{1}{2}}$ .

$\frac{v^{2}}{{(2 v - 5)}^{\frac{1}{2}}} + 2 \cdot {(2 v - 5)}^{\frac{1}{2}} v$

Step 17

Combine $\frac{v^{2}}{{(2 v - 5)}^{\frac{1}{2}}}$ and $2 {(2 v - 5)}^{\frac{1}{2}} v$ using a common denominator.

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Step 17.1

Move ${(2 v - 5)}^{\frac{1}{2}}$ .

$\frac{v^{2}}{{(2 v - 5)}^{\frac{1}{2}}} + 2 v {(2 v - 5)}^{\frac{1}{2}}$

Step 17.2

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

$\frac{v^{2}}{{(2 v - 5)}^{\frac{1}{2}}} + \frac{2 v {(2 v - 5)}^{\frac{1}{2}} {(2 v - 5)}^{\frac{1}{2}}}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 17.3

Combine the numerators over the common denominator.

$\frac{v^{2} + 2 v {(2 v - 5)}^{\frac{1}{2}} {(2 v - 5)}^{\frac{1}{2}}}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 18

Multiply ${(2 v - 5)}^{\frac{1}{2}}$ by ${(2 v - 5)}^{\frac{1}{2}}$ by adding the exponents.

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Step 18.1

Move ${(2 v - 5)}^{\frac{1}{2}}$ .

$\frac{v^{2} + 2 v ({(2 v - 5)}^{\frac{1}{2}} {(2 v - 5)}^{\frac{1}{2}})}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 18.2

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

$\frac{v^{2} + 2 v {(2 v - 5)}^{\frac{1}{2} + \frac{1}{2}}}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 18.3

Combine the numerators over the common denominator.

$\frac{v^{2} + 2 v {(2 v - 5)}^{\frac{1 + 1}{2}}}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 18.4

Add $1$ and $1$ .

$\frac{v^{2} + 2 v {(2 v - 5)}^{\frac{2}{2}}}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 18.5

Divide $2$ by $2$ .

$\frac{v^{2} + 2 v {(2 v - 5)}^{1}}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 19

Simplify $2 v {(2 v - 5)}^{1}$ .

$\frac{v^{2} + 2 v (2 v - 5)}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20

Simplify.

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Step 20.1

Apply the distributive property.

$\frac{v^{2} + 2 v (2 v) + 2 v \cdot - 5}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.2

Simplify the numerator.

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Step 20.2.1

Simplify each term.

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Step 20.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{v^{2} + 2 \cdot 2 v \cdot v + 2 v \cdot - 5}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.2.1.2

Multiply $v$ by $v$ by adding the exponents.

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Step 20.2.1.2.1

Move $v$ .

$\frac{v^{2} + 2 \cdot 2 (v \cdot v) + 2 v \cdot - 5}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.2.1.2.2

Multiply $v$ by $v$ .

$\frac{v^{2} + 2 \cdot 2 v^{2} + 2 v \cdot - 5}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.2.1.3

Multiply $2$ by $2$ .

$\frac{v^{2} + 4 v^{2} + 2 v \cdot - 5}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.2.1.4

Multiply $- 5$ by $2$ .

$\frac{v^{2} + 4 v^{2} - 10 v}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.2.2

Add $v^{2}$ and $4 v^{2}$ .

$\frac{5 v^{2} - 10 v}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.3

Factor $5 v$ out of $5 v^{2} - 10 v$ .

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Step 20.3.1

Factor $5 v$ out of $5 v^{2}$ .

$\frac{5 v (v) - 10 v}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.3.2

Factor $5 v$ out of $- 10 v$ .

$\frac{5 v (v) + 5 v (- 2)}{{(2 v - 5)}^{\frac{1}{2}}}$

Step 20.3.3

Factor $5 v$ out of $5 v (v) + 5 v (- 2)$ .

$\frac{5 v (v - 2)}{{(2 v - 5)}^{\frac{1}{2}}}$