Calculus Examples

$\frac{y^{4}}{8} + \frac{1}{4 y^{2}}$

Step 1

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

$\frac{d}{d y} [\frac{y^{4}}{8}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2

Evaluate $\frac{d}{d y} [\frac{y^{4}}{8}]$ .

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Step 2.1

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

$\frac{1}{8} \frac{d}{d y} [y^{4}] + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Combine $\frac{4}{8}$ and $y^{3}$ .

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

Step 2.5

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

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Step 2.5.1

Factor $4$ out of $4 y^{3}$ .

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

Step 2.5.2

Cancel the common factors.

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Step 2.5.2.1

Factor $4$ out of $8$ .

$\frac{4 y^{3}}{4 \cdot 2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.5.2.2

Cancel the common factor.

$\frac{4 y^{3}}{4 \cdot 2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 2.5.2.3

Rewrite the expression.

$\frac{y^{3}}{2} + \frac{d}{d y} [\frac{1}{4 y^{2}}]$

Step 3

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

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Step 3.1

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

$\frac{y^{3}}{2} + \frac{1}{4} \frac{d}{d y} [\frac{1}{y^{2}}]$

Step 3.2

Rewrite $\frac{1}{y^{2}}$ as ${(y^{2})}^{- 1}$ .

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

Step 3.3

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

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Step 3.3.1

To apply the Chain Rule, set $u$ as $y^{2}$ .

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

Step 3.3.2

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

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

Step 3.3.3

Replace all occurrences of $u$ with $y^{2}$ .

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

Step 3.4

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

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

Step 3.5

Multiply the exponents in ${(y^{2})}^{- 2}$ .

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Step 3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.5.2

Multiply $2$ by $- 2$ .

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

Step 3.6

Multiply $2$ by $- 1$ .

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

Step 3.7

Raise $y$ to the power of $1$ .

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

Step 3.8

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

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

Step 3.9

Subtract $4$ from $1$ .

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

Step 3.10

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

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

Step 3.11

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

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

Step 3.12

Move $y^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.13

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

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Step 3.13.1

Factor $2$ out of $- 2$ .

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

Step 3.13.2

Cancel the common factors.

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Step 3.13.2.1

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

$\frac{y^{3}}{2} + \frac{2 (- 1)}{2 (2 y^{3})}$

Step 3.13.2.2

Cancel the common factor.

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

Step 3.13.2.3

Rewrite the expression.

$\frac{y^{3}}{2} + \frac{- 1}{2 y^{3}}$

Step 3.14

Move the negative in front of the fraction.

$\frac{y^{3}}{2} - \frac{1}{2 y^{3}}$