Calculus Examples

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

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \frac{1}{y^{2}} - \frac{2}{y^{4}}$ and $g (x) = y + 8 y^{3}$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) \frac{d}{d x} [y + 8 y^{3}] + (y + 8 y^{3}) \frac{d}{d x} [\frac{1}{y^{2}} - \frac{2}{y^{4}}]$

Step 2

Differentiate.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $y + 8 y^{3}$ with respect to $x$ is $\frac{d}{d x} [y] + \frac{d}{d x} [8 y^{3}]$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) (\frac{d}{d x} [y] + \frac{d}{d x} [8 y^{3}]) + (y + 8 y^{3}) \frac{d}{d x} [\frac{1}{y^{2}} - \frac{2}{y^{4}}]$

Step 2.2

Since $y$ is constant with respect to $x$ , the derivative of $y$ with respect to $x$ is $0$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) (0 + \frac{d}{d x} [8 y^{3}]) + (y + 8 y^{3}) \frac{d}{d x} [\frac{1}{y^{2}} - \frac{2}{y^{4}}]$

Step 2.3

Since $8 y^{3}$ is constant with respect to $x$ , the derivative of $8 y^{3}$ with respect to $x$ is $0$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) (0 + 0) + (y + 8 y^{3}) \frac{d}{d x} [\frac{1}{y^{2}} - \frac{2}{y^{4}}]$

Step 2.4

Add $0$ and $0$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) \cdot 0 + (y + 8 y^{3}) \frac{d}{d x} [\frac{1}{y^{2}} - \frac{2}{y^{4}}]$

Step 2.5

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

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) \cdot 0 + (y + 8 y^{3}) (\frac{d}{d x} [\frac{1}{y^{2}}] + \frac{d}{d x} [- \frac{2}{y^{4}}])$

Step 2.6

Since $\frac{1}{y^{2}}$ is constant with respect to $x$ , the derivative of $\frac{1}{y^{2}}$ with respect to $x$ is $0$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) \cdot 0 + (y + 8 y^{3}) (0 + \frac{d}{d x} [- \frac{2}{y^{4}}])$

Step 2.7

Since $- \frac{2}{y^{4}}$ is constant with respect to $x$ , the derivative of $- \frac{2}{y^{4}}$ with respect to $x$ is $0$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) \cdot 0 + (y + 8 y^{3}) (0 + 0)$

Step 2.8

Add $0$ and $0$ .

$(\frac{1}{y^{2}} - \frac{2}{y^{4}}) \cdot 0 + (y + 8 y^{3}) \cdot 0$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Apply the distributive property.

$\frac{1}{y^{2}} \cdot 0 - \frac{2}{y^{4}} \cdot 0 + (y + 8 y^{3}) \cdot 0$

Step 3.2

Apply the distributive property.

$\frac{1}{y^{2}} \cdot 0 - \frac{2}{y^{4}} \cdot 0 + y \cdot 0 + 8 y^{3} \cdot 0$

Step 3.3

Combine terms.

Tap for more steps...

Step 3.3.1

Multiply $\frac{1}{y^{2}}$ by $0$ .

$0 - \frac{2}{y^{4}} \cdot 0 + y \cdot 0 + 8 y^{3} \cdot 0$

Step 3.3.2

Multiply $0$ by $- 1$ .

$0 + 0 \frac{2}{y^{4}} + y \cdot 0 + 8 y^{3} \cdot 0$

Step 3.3.3

Multiply $0$ by $\frac{2}{y^{4}}$ .

$0 + 0 + y \cdot 0 + 8 y^{3} \cdot 0$

Step 3.3.4

Add $0$ and $0$ .

$0 + y \cdot 0 + 8 y^{3} \cdot 0$

Step 3.3.5

Multiply $y$ by $0$ .

$0 + 0 + 8 y^{3} \cdot 0$

Step 3.3.6

Multiply $0$ by $8$ .

$0 + 0 + 0 y^{3}$

Step 3.3.7

Multiply $0$ by $y^{3}$ .

$0 + 0 + 0$

Step 3.3.8

Add $0$ and $0$ .

$0 + 0$

Step 3.3.9

Add $0$ and $0$ .

$0$