Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = y$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

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

Step 3.2.2.2

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

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

Step 3.2.2.3

Replace all occurrences of $u_{1}$ with $y$ .

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

Step 3.2.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

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

Step 3.2.4

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

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.2.7

Move $4$ to the left of $y^{3}$ .

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

Step 3.2.8

Cancel the common factor of $4$ .

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Step 3.2.8.1

Cancel the common factor.

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

Step 3.2.8.2

Divide $y^{3} y'$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

Since $\frac{1}{8}$ is constant with respect to $x$ , the derivative of $\frac{1}{8 y^{2}}$ with respect to $x$ is $\frac{1}{8} \frac{d}{d x} [\frac{1}{y^{2}}]$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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Step 3.3.4.1

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

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

Step 3.3.4.2

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

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

Step 3.3.4.3

Replace all occurrences of $u_{2}$ with $y$ .

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

Step 3.3.5

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

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

Step 3.3.6

Multiply $y^{2}$ by $0$ .

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

Step 3.3.7

Multiply $- 1$ by $1$ .

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

Step 3.3.8

Multiply $2$ by $- 1$ .

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

Step 3.3.9

Subtract $2 y y'$ from $0$ .

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

Step 3.3.10

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

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Step 3.3.10.1

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

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

Step 3.3.10.2

Multiply $2$ by $2$ .

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

Step 3.3.11

Cancel the common factor of $y$ and $y^{4}$ .

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Step 3.3.11.1

Factor $y$ out of $- 2 y y'$ .

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

Step 3.3.11.2

Cancel the common factors.

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Step 3.3.11.2.1

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

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

Step 3.3.11.2.2

Cancel the common factor.

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

Step 3.3.11.2.3

Rewrite the expression.

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

Step 3.3.12

Move the negative in front of the fraction.

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

Step 3.3.13

Multiply $\frac{1}{8}$ by $\frac{2 y'}{y^{3}}$ .

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

Step 3.3.14

Cancel the common factor of $2$ and $8$ .

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Step 3.3.14.1

Factor $2$ out of $2 y'$ .

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

Step 3.3.14.2

Cancel the common factors.

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Step 3.3.14.2.1

Factor $2$ out of $8 y^{3}$ .

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

Step 3.3.14.2.2

Cancel the common factor.

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

Step 3.3.14.2.3

Rewrite the expression.

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

Step 4

Reform the equation by setting the left side equal to the right side.

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

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $y^{3} y' - \frac{y'}{4 y^{3}} = 1$ .

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

Step 5.2

Find the LCD of the terms in the equation.

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Step 5.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, 4 y^{3}, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$4 y^{3}$

Step 5.3

Multiply each term in $y^{3} y' - \frac{y'}{4 y^{3}} = 1$ by $4 y^{3}$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $y^{3} y' - \frac{y'}{4 y^{3}} = 1$ by $4 y^{3}$ .

$y^{3} y' (4 y^{3}) - \frac{y'}{4 y^{3}} (4 y^{3}) = 1 (4 y^{3})$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Rewrite using the commutative property of multiplication.

$4 y^{3} y' y^{3} - \frac{y'}{4 y^{3}} (4 y^{3}) = 1 (4 y^{3})$

Step 5.3.2.1.2

Multiply $y^{3}$ by $y^{3}$ by adding the exponents.

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Step 5.3.2.1.2.1

Move $y^{3}$ .

$4 (y^{3} y^{3}) y' - \frac{y'}{4 y^{3}} (4 y^{3}) = 1 (4 y^{3})$

Step 5.3.2.1.2.2

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

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

Step 5.3.2.1.2.3

Add $3$ and $3$ .

$4 y^{6} y' - \frac{y'}{4 y^{3}} (4 y^{3}) = 1 (4 y^{3})$

Step 5.3.2.1.3

Cancel the common factor of $4 y^{3}$ .

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Step 5.3.2.1.3.1

Move the leading negative in $- \frac{y'}{4 y^{3}}$ into the numerator.

$4 y^{6} y' + \frac{- y'}{4 y^{3}} (4 y^{3}) = 1 (4 y^{3})$

Step 5.3.2.1.3.2

Cancel the common factor.

$4 y^{6} y' + \frac{- y'}{4 y^{3}} (4 y^{3}) = 1 (4 y^{3})$

Step 5.3.2.1.3.3

Rewrite the expression.

$4 y^{6} y' - y' = 1 (4 y^{3})$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $4 y^{3}$ by $1$ .

$4 y^{6} y' - y' = 4 y^{3}$

Step 5.4

Solve the equation.

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Step 5.4.1

Factor $y'$ out of $4 y^{6} y' - y'$ .

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Step 5.4.1.1

Factor $y'$ out of $4 y^{6} y'$ .

$y' (4 y^{6}) - y' = 4 y^{3}$

Step 5.4.1.2

Factor $y'$ out of $- y'$ .

$y' (4 y^{6}) + y' \cdot - 1 = 4 y^{3}$

Step 5.4.1.3

Factor $y'$ out of $y' (4 y^{6}) + y' \cdot - 1$ .

$y' (4 y^{6} - 1) = 4 y^{3}$

Step 5.4.2

Rewrite $4 y^{6}$ as ${(2 y^{3})}^{2}$ .

$y' ({(2 y^{3})}^{2} - 1) = 4 y^{3}$

Step 5.4.3

Rewrite $1$ as $1^{2}$ .

$y' ({(2 y^{3})}^{2} - 1^{2}) = 4 y^{3}$

Step 5.4.4

Factor.

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Step 5.4.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 y^{3}$ and $b = 1$ .

$y' ((2 y^{3} + 1) (2 y^{3} - 1)) = 4 y^{3}$

Step 5.4.4.2

Remove unnecessary parentheses.

$y' (2 y^{3} + 1) (2 y^{3} - 1) = 4 y^{3}$

Step 5.4.5

Divide each term in $y' (2 y^{3} + 1) (2 y^{3} - 1) = 4 y^{3}$ by $(2 y^{3} + 1) (2 y^{3} - 1)$ and simplify.

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Step 5.4.5.1

Divide each term in $y' (2 y^{3} + 1) (2 y^{3} - 1) = 4 y^{3}$ by $(2 y^{3} + 1) (2 y^{3} - 1)$ .

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

Step 5.4.5.2

Simplify the left side.

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Step 5.4.5.2.1

Cancel the common factor of $2 y^{3} + 1$ .

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Step 5.4.5.2.1.1

Cancel the common factor.

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

Step 5.4.5.2.1.2

Rewrite the expression.

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

Step 5.4.5.2.2

Cancel the common factor of $2 y^{3} - 1$ .

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Step 5.4.5.2.2.1

Cancel the common factor.

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

Step 5.4.5.2.2.2

Divide $y'$ by $1$ .

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

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

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