Calculus Examples

$g (z) = 4 z {(2 z^{2} - 6 z + 1)}^{4}$

Step 1

Since $4$ is constant with respect to $z$ , the derivative of $4 z {(2 z^{2} - 6 z + 1)}^{4}$ with respect to $z$ is $4 \frac{d}{d z} [z {(2 z^{2} - 6 z + 1)}^{4}]$ .

$4 \frac{d}{d z} [z {(2 z^{2} - 6 z + 1)}^{4}]$

Step 2

Differentiate using the Product Rule which states that $\frac{d}{d z} [f (z) g (z)]$ is $f (z) \frac{d}{d z} [g (z)] + g (z) \frac{d}{d z} [f (z)]$ where $f (z) = z$ and $g (z) = {(2 z^{2} - 6 z + 1)}^{4}$ .

$4 (z \frac{d}{d z} [{(2 z^{2} - 6 z + 1)}^{4}] + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d z} [f (g (z))]$ is $f' (g (z)) g' (z)$ where $f (z) = z^{4}$ and $g (z) = 2 z^{2} - 6 z + 1$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $2 z^{2} - 6 z + 1$ .

$4 (z (\frac{d}{d u} [u^{4}] \frac{d}{d z} [2 z^{2} - 6 z + 1]) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 3.2

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

$4 (z (4 u^{3} \frac{d}{d z} [2 z^{2} - 6 z + 1]) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 3.3

Replace all occurrences of $u$ with $2 z^{2} - 6 z + 1$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} \frac{d}{d z} [2 z^{2} - 6 z + 1]) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4

Differentiate.

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Step 4.1

By the Sum Rule, the derivative of $2 z^{2} - 6 z + 1$ with respect to $z$ is $\frac{d}{d z} [2 z^{2}] + \frac{d}{d z} [- 6 z] + \frac{d}{d z} [1]$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (\frac{d}{d z} [2 z^{2}] + \frac{d}{d z} [- 6 z] + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.2

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

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (2 \frac{d}{d z} [z^{2}] + \frac{d}{d z} [- 6 z] + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.3

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

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (2 (2 z) + \frac{d}{d z} [- 6 z] + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.4

Multiply $2$ by $2$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z + \frac{d}{d z} [- 6 z] + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.5

Since $- 6$ is constant with respect to $z$ , the derivative of $- 6 z$ with respect to $z$ is $- 6 \frac{d}{d z} [z]$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6 \frac{d}{d z} [z] + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.6

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

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6 \cdot 1 + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.7

Multiply $- 6$ by $1$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6 + \frac{d}{d z} [1])) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.8

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

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6 + 0)) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.9

Add $4 z - 6$ and $0$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6)) + {(2 z^{2} - 6 z + 1)}^{4} \frac{d}{d z} [z])$

Step 4.10

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

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6)) + {(2 z^{2} - 6 z + 1)}^{4} \cdot 1)$

Step 4.11

Multiply ${(2 z^{2} - 6 z + 1)}^{4}$ by $1$ .

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6)) + {(2 z^{2} - 6 z + 1)}^{4})$

Step 5

Simplify.

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Step 5.1

Apply the distributive property.

$4 (z (4 {(2 z^{2} - 6 z + 1)}^{3} (4 z - 6))) + 4 {(2 z^{2} - 6 z + 1)}^{4}$

Step 5.2

Multiply $4$ by $4$ .

$16 (z ({(2 z^{2} - 6 z + 1)}^{3} (4 z - 6))) + 4 {(2 z^{2} - 6 z + 1)}^{4}$

Step 5.3

Factor $4 {(2 z^{2} - 6 z + 1)}^{3}$ out of $16 z ({(2 z^{2} - 6 z + 1)}^{3} (4 z - 6)) + 4 {(2 z^{2} - 6 z + 1)}^{4}$ .

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Step 5.3.1

Factor $4 {(2 z^{2} - 6 z + 1)}^{3}$ out of $16 z ({(2 z^{2} - 6 z + 1)}^{3} (4 z - 6))$ .

$4 {(2 z^{2} - 6 z + 1)}^{3} (4 z (4 z - 6)) + 4 {(2 z^{2} - 6 z + 1)}^{4}$

Step 5.3.2

Factor $4 {(2 z^{2} - 6 z + 1)}^{3}$ out of $4 {(2 z^{2} - 6 z + 1)}^{4}$ .

$4 {(2 z^{2} - 6 z + 1)}^{3} (4 z (4 z - 6)) + 4 {(2 z^{2} - 6 z + 1)}^{3} (2 z^{2} - 6 z + 1)$

Step 5.3.3

Factor $4 {(2 z^{2} - 6 z + 1)}^{3}$ out of $4 {(2 z^{2} - 6 z + 1)}^{3} (4 z (4 z - 6)) + 4 {(2 z^{2} - 6 z + 1)}^{3} (2 z^{2} - 6 z + 1)$ .

$4 {(2 z^{2} - 6 z + 1)}^{3} (4 z (4 z - 6) + 2 z^{2} - 6 z + 1)$