Calculus Examples

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

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

Step 1

By the Sum Rule, the derivative of

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

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

y

$y$ is

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

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

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

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

Step 2

Evaluate

\frac{d}{d y} [\frac{y^{4}}{4}]

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

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Step 2.1

Since

\frac{1}{4}

$\frac{1}{4}$ is constant with respect to

y

$y$ , the derivative of

\frac{y^{4}}{4}

$\frac{y^{4}}{4}$ with respect to

y

$y$ is

\frac{1}{4} \frac{d}{d y} [y^{4}]

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

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

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

Step 2.2

Differentiate using the Power Rule which states that

\frac{d}{d y} [y^{n}]

$\frac{d}{d y} [y^{n}]$ is

n y^{n - 1}

$n y^{n - 1}$ where

n = 4

$n = 4$ .

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

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

Step 2.3

Combine

4

$4$ and

\frac{1}{4}

$\frac{1}{4}$ .

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

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

Step 2.4

Combine

\frac{4}{4}

$\frac{4}{4}$ and

y^{3}

$y^{3}$ .

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

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

Step 2.5

Cancel the common factor of

4

$4$ .

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Step 2.5.1

Cancel the common factor.

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

Step 2.5.2

Divide

$y^{3}$ by

$1$ .

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

Step 3

Evaluate

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

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Step 3.1

Since

$\frac{1}{8}$ is constant with respect to

$y$ , the derivative of

$\frac{1}{8 y^{2}}$ with respect to

$y$ is

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

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

Step 3.2

Rewrite

$\frac{1}{y^{2}}$ as

${(y^{2})}^{- 1}$ .

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

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

$y^{2}$ .

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

Step 3.3.2

Differentiate using the Power Rule which states that

$\frac{d}{d u} [u^{n}]$ is

$n u^{n - 1}$ where

$n = - 1$ .

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

Step 3.3.3

Replace all occurrences of

$u$ with

$y^{2}$ .

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

Step 3.4

Differentiate using the Power Rule which states that

$\frac{d}{d y} [y^{n}]$ is

$n y^{n - 1}$ where

$n = 2$ .

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

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,

${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.5.2

Multiply

$2$ by

$- 2$ .

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

Step 3.6

Multiply

$2$ by

$- 1$ .

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

Step 3.7

Raise

$y$ to the power of

$1$ .

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

Step 3.8

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.9

Subtract

$4$ from

$1$ .

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

Step 3.10

Combine

$- 2$ and

$\frac{1}{8}$ .

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

Step 3.11

Combine

$\frac{- 2}{8}$ and

$y^{- 3}$ .

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

Step 3.12

Move

$y^{- 3}$ to the denominator using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.13

Cancel the common factor of

$- 2$ and

$8$ .

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Step 3.13.1

Factor

$2$ out of

$- 2$ .

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

Step 3.13.2

Cancel the common factors.

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Step 3.13.2.1

Factor

$2$ out of

$8 y^{3}$ .

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

Step 3.13.2.2

Cancel the common factor.

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

Step 3.13.2.3

Rewrite the expression.

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

Step 3.14

Move the negative in front of the fraction.

$y^{3} - \frac{1}{4 y^{3}}$