Calculus Examples

$f (z) = - \frac{1}{4} z^{8} + \frac{1}{2} z^{4} - 2^{3}$

Step 1

Differentiate using the Sum Rule.

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Step 1.1

Raise $2$ to the power of $3$ .

$\frac{d}{d z} [- \frac{1}{4} z^{8} + \frac{1}{2} z^{4} - 1 \cdot 8]$

Step 1.2

By the Sum Rule, the derivative of $- \frac{1}{4} z^{8} + \frac{1}{2} z^{4} - 1 \cdot 8$ with respect to $z$ is $\frac{d}{d z} [- \frac{1}{4} z^{8}] + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$ .

$\frac{d}{d z} [- \frac{1}{4} z^{8}] + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2

Evaluate $\frac{d}{d z} [- \frac{1}{4} z^{8}]$ .

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Step 2.1

Since $- \frac{1}{4}$ is constant with respect to $z$ , the derivative of $- \frac{1}{4} z^{8}$ with respect to $z$ is $- \frac{1}{4} \frac{d}{d z} [z^{8}]$ .

$- \frac{1}{4} \frac{d}{d z} [z^{8}] + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.2

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

$- \frac{1}{4} (8 z^{7}) + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.3

Multiply $8$ by $- 1$ .

$- 8 (\frac{1}{4}) z^{7} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.4

Combine $- 8$ and $\frac{1}{4}$ .

$\frac{- 8}{4} z^{7} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.5

Combine $\frac{- 8}{4}$ and $z^{7}$ .

$\frac{- 8 z^{7}}{4} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.6

Cancel the common factor of $- 8$ and $4$ .

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Step 2.6.1

Factor $4$ out of $- 8 z^{7}$ .

$\frac{4 (- 2 z^{7})}{4} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.6.2

Cancel the common factors.

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Step 2.6.2.1

Factor $4$ out of $4$ .

$\frac{4 (- 2 z^{7})}{4 (1)} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.6.2.2

Cancel the common factor.

$\frac{4 (- 2 z^{7})}{4 \cdot 1} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.6.2.3

Rewrite the expression.

$\frac{- 2 z^{7}}{1} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 2.6.2.4

Divide $- 2 z^{7}$ by $1$ .

$- 2 z^{7} + \frac{d}{d z} [\frac{1}{2} z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 3

Evaluate $\frac{d}{d z} [\frac{1}{2} z^{4}]$ .

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Step 3.1

Since $\frac{1}{2}$ is constant with respect to $z$ , the derivative of $\frac{1}{2} z^{4}$ with respect to $z$ is $\frac{1}{2} \frac{d}{d z} [z^{4}]$ .

$- 2 z^{7} + \frac{1}{2} \frac{d}{d z} [z^{4}] + \frac{d}{d z} [- 1 \cdot 8]$

Step 3.2

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

$- 2 z^{7} + \frac{1}{2} (4 z^{3}) + \frac{d}{d z} [- 1 \cdot 8]$

Step 3.3

Combine $4$ and $\frac{1}{2}$ .

$- 2 z^{7} + \frac{4}{2} z^{3} + \frac{d}{d z} [- 1 \cdot 8]$

Step 3.4

Combine $\frac{4}{2}$ and $z^{3}$ .

$- 2 z^{7} + \frac{4 z^{3}}{2} + \frac{d}{d z} [- 1 \cdot 8]$

Step 3.5

Cancel the common factor of $4$ and $2$ .

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Step 3.5.1

Factor $2$ out of $4 z^{3}$ .

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

Step 3.5.2

Cancel the common factors.

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Step 3.5.2.1

Factor $2$ out of $2$ .

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

Step 3.5.2.2

Cancel the common factor.

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

Step 3.5.2.3

Rewrite the expression.

$- 2 z^{7} + \frac{2 z^{3}}{1} + \frac{d}{d z} [- 1 \cdot 8]$

Step 3.5.2.4

Divide $2 z^{3}$ by $1$ .

$- 2 z^{7} + 2 z^{3} + \frac{d}{d z} [- 1 \cdot 8]$

Step 4

Evaluate $\frac{d}{d z} [- 1 \cdot 8]$ .

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Step 4.1

Multiply $- 1$ by $8$ .

$- 2 z^{7} + 2 z^{3} + \frac{d}{d z} [- 8]$

Step 4.2

Since $- 8$ is constant with respect to $z$ , the derivative of $- 8$ with respect to $z$ is $0$ .

$- 2 z^{7} + 2 z^{3} + 0$

Step 5

Add $- 2 z^{7} + 2 z^{3}$ and $0$ .

$- 2 z^{7} + 2 z^{3}$