Calculus Examples

$y = \frac{2 - v^{2}}{2 + v^{2}}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d v} [\frac{f (v)}{g (v)}]$ is $\frac{g (v) \frac{d}{d v} [f (v)] - f (v) \frac{d}{d v} [g (v)]}{{g (v)}^{2}}$ where $f (v) = 2 - v^{2}$ and $g (v) = 2 + v^{2}$ .

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

Since $2$ is constant with respect to $v$ , the derivative of $2$ with respect to $v$ is $0$ .

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

Step 3

Evaluate $\frac{d}{d v} [- v^{2}]$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Multiply $2$ by $- 1$ .

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

Step 4

Differentiate.

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Step 4.1

Subtract $2 v$ from $0$ .

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

Step 4.2

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

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

Step 4.3

Since $2$ is constant with respect to $v$ , the derivative of $2$ with respect to $v$ is $0$ .

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

Step 4.4

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

$\frac{(2 + v^{2}) (- 2 v) - (2 - v^{2}) (0 + 2 v)}{{(2 + v^{2})}^{2}}$

Step 4.5

Add $0$ and $2 v$ .

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

Step 5

Simplify.

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Step 5.1

Apply the distributive property.

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

Step 5.2

Apply the distributive property.

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

Step 5.3

Apply the distributive property.

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

Step 5.4

Simplify the numerator.

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Step 5.4.1

Simplify each term.

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Step 5.4.1.1

Multiply $- 2$ by $2$ .

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

Step 5.4.1.2

Rewrite using the commutative property of multiplication.

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

Step 5.4.1.3

Multiply $v^{2}$ by $v$ by adding the exponents.

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Step 5.4.1.3.1

Move $v$ .

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

Step 5.4.1.3.2

Multiply $v$ by $v^{2}$ .

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Step 5.4.1.3.2.1

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

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

Step 5.4.1.3.2.2

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

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

Step 5.4.1.3.3

Add $1$ and $2$ .

$\frac{- 4 v - 2 v^{3} - 1 \cdot 2 (2 v) - - v^{2} (2 v)}{{(2 + v^{2})}^{2}}$

Step 5.4.1.4

Multiply $- 1$ by $2$ .

$\frac{- 4 v - 2 v^{3} - 2 (2 v) - - v^{2} (2 v)}{{(2 + v^{2})}^{2}}$

Step 5.4.1.5

Multiply $2$ by $- 2$ .

$\frac{- 4 v - 2 v^{3} - 4 v - - v^{2} (2 v)}{{(2 + v^{2})}^{2}}$

Step 5.4.1.6

Multiply $v^{2}$ by $v$ by adding the exponents.

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Step 5.4.1.6.1

Move $v$ .

$\frac{- 4 v - 2 v^{3} - 4 v - - (v \cdot v^{2}) \cdot 2}{{(2 + v^{2})}^{2}}$

Step 5.4.1.6.2

Multiply $v$ by $v^{2}$ .

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Step 5.4.1.6.2.1

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

$\frac{- 4 v - 2 v^{3} - 4 v - - (v^{1} v^{2}) \cdot 2}{{(2 + v^{2})}^{2}}$

Step 5.4.1.6.2.2

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

$\frac{- 4 v - 2 v^{3} - 4 v - - v^{1 + 2} \cdot 2}{{(2 + v^{2})}^{2}}$

Step 5.4.1.6.3

Add $1$ and $2$ .

$\frac{- 4 v - 2 v^{3} - 4 v - - v^{3} \cdot 2}{{(2 + v^{2})}^{2}}$

Step 5.4.1.7

Multiply $- - v^{3}$ .

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Step 5.4.1.7.1

Multiply $- 1$ by $- 1$ .

$\frac{- 4 v - 2 v^{3} - 4 v + 1 v^{3} \cdot 2}{{(2 + v^{2})}^{2}}$

Step 5.4.1.7.2

Multiply $v^{3}$ by $1$ .

$\frac{- 4 v - 2 v^{3} - 4 v + v^{3} \cdot 2}{{(2 + v^{2})}^{2}}$

Step 5.4.1.8

Move $2$ to the left of $v^{3}$ .

$\frac{- 4 v - 2 v^{3} - 4 v + 2 v^{3}}{{(2 + v^{2})}^{2}}$

Step 5.4.2

Combine the opposite terms in $- 4 v - 2 v^{3} - 4 v + 2 v^{3}$ .

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Step 5.4.2.1

Add $- 2 v^{3}$ and $2 v^{3}$ .

$\frac{- 4 v - 4 v + 0}{{(2 + v^{2})}^{2}}$

Step 5.4.2.2

Add $- 4 v - 4 v$ and $0$ .

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

Step 5.4.3

Subtract $4 v$ from $- 4 v$ .

$\frac{- 8 v}{{(2 + v^{2})}^{2}}$

Step 5.5

Move the negative in front of the fraction.

$- \frac{8 v}{{(2 + v^{2})}^{2}}$