Calculus Examples

$y = {(\frac{1}{w^{3} - 1})}^{8}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d w} [f (g (w))]$ is $f' (g (w)) g' (w)$ where $f (w) = w^{8}$ and $g (w) = \frac{1}{w^{3} - 1}$ .

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Step 1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{8}] \frac{d}{d w} [\frac{1}{w^{3} - 1}]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 8$ .

$8 {u_{1}}^{7} \frac{d}{d w} [\frac{1}{w^{3} - 1}]$

Step 1.3

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

$8 {(\frac{1}{w^{3} - 1})}^{7} \frac{d}{d w} [\frac{1}{w^{3} - 1}]$

Step 2

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

$8 {(\frac{1}{w^{3} - 1})}^{7} \frac{d}{d w} [{(w^{3} - 1)}^{- 1}]$

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $w^{3} - 1$ .

$8 {(\frac{1}{w^{3} - 1})}^{7} (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d w} [w^{3} - 1])$

Step 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 $n = - 1$ .

$8 {(\frac{1}{w^{3} - 1})}^{7} (- {u_{2}}^{- 2} \frac{d}{d w} [w^{3} - 1])$

Step 3.3

Replace all occurrences of $u_{2}$ with $w^{3} - 1$ .

$8 {(\frac{1}{w^{3} - 1})}^{7} (- {(w^{3} - 1)}^{- 2} \frac{d}{d w} [w^{3} - 1])$

Step 4

Differentiate.

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Step 4.1

Multiply $- 1$ by $8$ .

$- 8 {(\frac{1}{w^{3} - 1})}^{7} ({(w^{3} - 1)}^{- 2} \frac{d}{d w} [w^{3} - 1])$

Step 4.2

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

$- 8 {(\frac{1}{w^{3} - 1})}^{7} ({(w^{3} - 1)}^{- 2} (\frac{d}{d w} [w^{3}] + \frac{d}{d w} [- 1]))$

Step 4.3

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

$- 8 {(\frac{1}{w^{3} - 1})}^{7} ({(w^{3} - 1)}^{- 2} (3 w^{2} + \frac{d}{d w} [- 1]))$

Step 4.4

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

$- 8 {(\frac{1}{w^{3} - 1})}^{7} ({(w^{3} - 1)}^{- 2} (3 w^{2} + 0))$

Step 4.5

Simplify the expression.

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Step 4.5.1

Add $3 w^{2}$ and $0$ .

$- 8 {(\frac{1}{w^{3} - 1})}^{7} ({(w^{3} - 1)}^{- 2} (3 w^{2}))$

Step 4.5.2

Move $3$ to the left of ${(w^{3} - 1)}^{- 2}$ .

$- 8 {(\frac{1}{w^{3} - 1})}^{7} (3 \cdot {(w^{3} - 1)}^{- 2} w^{2})$

Step 4.5.3

Multiply $3$ by $- 8$ .

$- 24 {(\frac{1}{w^{3} - 1})}^{7} ({(w^{3} - 1)}^{- 2} w^{2})$

Step 5

Simplify.

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Step 5.1

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

$- 24 {(\frac{1}{w^{3} - 1})}^{7} (\frac{1}{{(w^{3} - 1)}^{2}} w^{2})$

Step 5.2

Apply the product rule to $\frac{1}{w^{3} - 1}$ .

$- 24 \frac{1^{7}}{{(w^{3} - 1)}^{7}} (\frac{1}{{(w^{3} - 1)}^{2}} w^{2})$

Step 5.3

Combine terms.

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Step 5.3.1

One to any power is one.

$- 24 \frac{1}{{(w^{3} - 1)}^{7}} (\frac{1}{{(w^{3} - 1)}^{2}} w^{2})$

Step 5.3.2

Combine $- 24$ and $\frac{1}{{(w^{3} - 1)}^{7}}$ .

$\frac{- 24}{{(w^{3} - 1)}^{7}} (\frac{1}{{(w^{3} - 1)}^{2}} w^{2})$

Step 5.3.3

Move the negative in front of the fraction.

$- \frac{24}{{(w^{3} - 1)}^{7}} (\frac{1}{{(w^{3} - 1)}^{2}} w^{2})$

Step 5.3.4

Combine $\frac{1}{{(w^{3} - 1)}^{2}}$ and $w^{2}$ .

$- \frac{24}{{(w^{3} - 1)}^{7}} \cdot \frac{w^{2}}{{(w^{3} - 1)}^{2}}$

Step 5.3.5

Multiply $\frac{w^{2}}{{(w^{3} - 1)}^{2}}$ by $\frac{24}{{(w^{3} - 1)}^{7}}$ .

$- \frac{w^{2} \cdot 24}{{(w^{3} - 1)}^{2} {(w^{3} - 1)}^{7}}$

Step 5.3.6

Multiply ${(w^{3} - 1)}^{2}$ by ${(w^{3} - 1)}^{7}$ by adding the exponents.

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Step 5.3.6.1

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

$- \frac{w^{2} \cdot 24}{{(w^{3} - 1)}^{2 + 7}}$

Step 5.3.6.2

Add $2$ and $7$ .

$- \frac{w^{2} \cdot 24}{{(w^{3} - 1)}^{9}}$

Step 5.3.7

Move $24$ to the left of $w^{2}$ .

$- \frac{24 w^{2}}{{(w^{3} - 1)}^{9}}$

Step 5.4

Simplify the denominator.

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Step 5.4.1

Rewrite $1$ as $1^{3}$ .

$- \frac{24 w^{2}}{{(w^{3} - 1^{3})}^{9}}$

Step 5.4.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = w$ and $b = 1$ .

$- \frac{24 w^{2}}{{((w - 1) (w^{2} + w \cdot 1 + 1^{2}))}^{9}}$

Step 5.4.3

Simplify.

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Step 5.4.3.1

Multiply $w$ by $1$ .

$- \frac{24 w^{2}}{{((w - 1) (w^{2} + w + 1^{2}))}^{9}}$

Step 5.4.3.2

One to any power is one.

$- \frac{24 w^{2}}{{((w - 1) (w^{2} + w + 1))}^{9}}$

Step 5.4.4

Apply the product rule to $(w - 1) (w^{2} + w + 1)$ .

$- \frac{24 w^{2}}{{(w - 1)}^{9} {(w^{2} + w + 1)}^{9}}$