Calculus Examples

${(v^{- 3} + 7 v^{- 2})}^{3}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d v} [f (g (v))]$ is $f' (g (v)) g' (v)$ where $f (v) = v^{3}$ and $g (v) = v^{- 3} + 7 v^{- 2}$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $v^{- 3} + 7 v^{- 2}$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d v} [v^{- 3} + 7 v^{- 2}]$

Step 1.2

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

$3 u^{2} \frac{d}{d v} [v^{- 3} + 7 v^{- 2}]$

Step 1.3

Replace all occurrences of $u$ with $v^{- 3} + 7 v^{- 2}$ .

$3 {(v^{- 3} + 7 v^{- 2})}^{2} \frac{d}{d v} [v^{- 3} + 7 v^{- 2}]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $v^{- 3} + 7 v^{- 2}$ with respect to $v$ is $\frac{d}{d v} [v^{- 3}] + \frac{d}{d v} [7 v^{- 2}]$ .

$3 {(v^{- 3} + 7 v^{- 2})}^{2} (\frac{d}{d v} [v^{- 3}] + \frac{d}{d v} [7 v^{- 2}])$

Step 2.2

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

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

Step 2.3

Since $7$ is constant with respect to $v$ , the derivative of $7 v^{- 2}$ with respect to $v$ is $7 \frac{d}{d v} [v^{- 2}]$ .

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

Step 2.4

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

$3 {(v^{- 3} + 7 v^{- 2})}^{2} (- 3 v^{- 4} + 7 (- 2 v^{- 3}))$

Step 2.5

Multiply $- 2$ by $7$ .

$3 {(v^{- 3} + 7 v^{- 2})}^{2} (- 3 v^{- 4} - 14 v^{- 3})$

Step 3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 {(\frac{1}{v^{3}} + 7 v^{- 2})}^{2} (- 3 v^{- 4} - 14 v^{- 3})$

Step 4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 {(\frac{1}{v^{3}} + 7 \frac{1}{v^{2}})}^{2} (- 3 v^{- 4} - 14 v^{- 3})$

Step 5

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 {(\frac{1}{v^{3}} + 7 \frac{1}{v^{2}})}^{2} (- 3 \frac{1}{v^{4}} - 14 v^{- 3})$

Step 6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 {(\frac{1}{v^{3}} + 7 \frac{1}{v^{2}})}^{2} (- 3 \frac{1}{v^{4}} - 14 \frac{1}{v^{3}})$

Step 7

Combine terms.

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Step 7.1

Combine $7$ and $\frac{1}{v^{2}}$ .

$3 {(\frac{1}{v^{3}} + \frac{7}{v^{2}})}^{2} (- 3 \frac{1}{v^{4}} - 14 \frac{1}{v^{3}})$

Step 7.2

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

$3 {(\frac{1}{v^{3}} + \frac{7}{v^{2}})}^{2} (\frac{- 3}{v^{4}} - 14 \frac{1}{v^{3}})$

Step 7.3

Move the negative in front of the fraction.

$3 {(\frac{1}{v^{3}} + \frac{7}{v^{2}})}^{2} (- \frac{3}{v^{4}} - 14 \frac{1}{v^{3}})$

Step 7.4

Combine $- 14$ and $\frac{1}{v^{3}}$ .

$3 {(\frac{1}{v^{3}} + \frac{7}{v^{2}})}^{2} (- \frac{3}{v^{4}} + \frac{- 14}{v^{3}})$

Step 7.5

Move the negative in front of the fraction.

$3 {(\frac{1}{v^{3}} + \frac{7}{v^{2}})}^{2} (- \frac{3}{v^{4}} - \frac{14}{v^{3}})$