Calculus Examples

$x = \frac{y^{3}}{27} + \frac{9}{4 y}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x) = \frac{d}{d x} (\frac{y^{3}}{27} + \frac{9}{4 y})$

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 $\frac{y^{3}}{27} + \frac{9}{4 y}$ with respect to $x$ is $\frac{d}{d x} [\frac{y^{3}}{27}] + \frac{d}{d x} [\frac{9}{4 y}]$ .

$\frac{d}{d x} [\frac{y^{3}}{27}] + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2

Evaluate $\frac{d}{d x} [\frac{y^{3}}{27}]$ .

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Step 3.2.1

Since $\frac{1}{27}$ is constant with respect to $x$ , the derivative of $\frac{y^{3}}{27}$ with respect to $x$ is $\frac{1}{27} \frac{d}{d x} [y^{3}]$ .

$\frac{1}{27} \frac{d}{d x} [y^{3}] + \frac{d}{d x} [\frac{9}{4 y}]$

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 $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u$ as $y$ .

$\frac{1}{27} (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$\frac{1}{27} (3 u^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.2.3

Replace all occurrences of $u$ with $y$ .

$\frac{1}{27} (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.3

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

$\frac{1}{27} (3 y^{2} y') + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.4

Combine $3$ and $\frac{1}{27}$ .

$\frac{3}{27} (y^{2} y') + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.5

Combine $y^{2}$ and $\frac{3}{27}$ .

$\frac{y^{2} \cdot 3}{27} y' + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.6

Combine $\frac{y^{2} \cdot 3}{27}$ and $y'$ .

$\frac{y^{2} \cdot 3 y'}{27} + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.7

Move $3$ to the left of $y^{2}$ .

$\frac{3 \cdot y^{2} y'}{27} + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.8

Cancel the common factor of $3$ and $27$ .

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Step 3.2.8.1

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

$\frac{3 (y^{2} y')}{27} + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.8.2

Cancel the common factors.

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Step 3.2.8.2.1

Factor $3$ out of $27$ .

$\frac{3 (y^{2} y')}{3 \cdot 9} + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.8.2.2

Cancel the common factor.

$\frac{3 (y^{2} y')}{3 \cdot 9} + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.2.8.2.3

Rewrite the expression.

$\frac{y^{2} y'}{9} + \frac{d}{d x} [\frac{9}{4 y}]$

Step 3.3

Evaluate $\frac{d}{d x} [\frac{9}{4 y}]$ .

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Step 3.3.1

Since $\frac{9}{4}$ is constant with respect to $x$ , the derivative of $\frac{9}{4 y}$ with respect to $x$ is $\frac{9}{4} \frac{d}{d x} [\frac{1}{y}]$ .

$\frac{y^{2} y'}{9} + \frac{9}{4} \frac{d}{d x} [\frac{1}{y}]$

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 $g (x) = y$ .

$\frac{y^{2} y'}{9} + \frac{9}{4} \cdot \frac{y \frac{d}{d x} [1] - 1 \cdot 1 \frac{d}{d x} [y]}{y^{2}}$

Step 3.3.3

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

$\frac{y^{2} y'}{9} + \frac{9}{4} \cdot \frac{y \cdot 0 - 1 \cdot 1 \frac{d}{d x} [y]}{y^{2}}$

Step 3.3.4

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

$\frac{y^{2} y'}{9} + \frac{9}{4} \cdot \frac{y \cdot 0 - 1 \cdot 1 y'}{y^{2}}$

Step 3.3.5

Multiply $y$ by $0$ .

$\frac{y^{2} y'}{9} + \frac{9}{4} \cdot \frac{0 - 1 \cdot 1 y'}{y^{2}}$

Step 3.3.6

Multiply $- 1$ by $1$ .

$\frac{y^{2} y'}{9} + \frac{9}{4} \cdot \frac{0 - y'}{y^{2}}$

Step 3.3.7

Subtract $y'$ from $0$ .

$\frac{y^{2} y'}{9} + \frac{9}{4} \cdot \frac{- y'}{y^{2}}$

Step 3.3.8

Move the negative in front of the fraction.

$\frac{y^{2} y'}{9} + \frac{9}{4} (- \frac{y'}{y^{2}})$

Step 3.3.9

Multiply $\frac{9}{4}$ by $\frac{y'}{y^{2}}$ .

$\frac{y^{2} y'}{9} - \frac{9 y'}{4 y^{2}}$

Step 4

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

$1 = \frac{y^{2} y'}{9} - \frac{9 y'}{4 y^{2}}$

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $\frac{y^{2} y'}{9} - \frac{9 y'}{4 y^{2}} = 1$ .

$\frac{y^{2} y'}{9} - \frac{9 y'}{4 y^{2}} = 1$

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.

$9, 4 y^{2}, 1$

Step 5.2.2

Since $9, 4 y^{2}, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $9, 4, 1$ then find LCM for the variable part $y^{2}$ .

Step 5.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 5.2.4

$9$ has factors of $3$ and $3$ .

$3 \cdot 3$

Step 5.2.5

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Step 5.2.6

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 5.2.7

The LCM of $9, 4, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 2 \cdot 3 \cdot 3$

Step 5.2.8

Multiply $2 \cdot 2 \cdot 3 \cdot 3$ .

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Step 5.2.8.1

Multiply $2$ by $2$ .

$4 \cdot 3 \cdot 3$

Step 5.2.8.2

Multiply $4$ by $3$ .

$12 \cdot 3$

Step 5.2.8.3

Multiply $12$ by $3$ .

$36$

Step 5.2.9

The factors for $y^{2}$ are $y \cdot y$ , which is $y$ multiplied by each other $2$ times.

$y^{2} = y \cdot y$

$y$ occurs $2$ times.

Step 5.2.10

The LCM of $y^{2}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$y \cdot y$

Step 5.2.11

Multiply $y$ by $y$ .

$y^{2}$

Step 5.2.12

The LCM for $9, 4 y^{2}, 1$ is the numeric part $36$ multiplied by the variable part.

$36 y^{2}$

Step 5.3

Multiply each term in $\frac{y^{2} y'}{9} - \frac{9 y'}{4 y^{2}} = 1$ by $36 y^{2}$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $\frac{y^{2} y'}{9} - \frac{9 y'}{4 y^{2}} = 1$ by $36 y^{2}$ .

$\frac{y^{2} y'}{9} (36 y^{2}) - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

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.

$36 \frac{y^{2} y'}{9} y^{2} - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.2

Cancel the common factor of $9$ .

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Step 5.3.2.1.2.1

Factor $9$ out of $36$ .

$9 (4) \frac{y^{2} y'}{9} y^{2} - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.2.2

Cancel the common factor.

$9 \cdot 4 \frac{y^{2} y'}{9} y^{2} - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.2.3

Rewrite the expression.

$4 (y^{2} y') y^{2} - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.3

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

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Step 5.3.2.1.3.1

Move $y^{2}$ .

$4 (y^{2} y^{2} y') - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.3.2

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

$4 (y^{2 + 2} y') - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.3.3

Add $2$ and $2$ .

$4 (y^{4} y') - \frac{9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.4

Cancel the common factor of $4 y^{2}$ .

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Step 5.3.2.1.4.1

Move the leading negative in $- \frac{9 y'}{4 y^{2}}$ into the numerator.

$4 y^{4} y' + \frac{- 9 y'}{4 y^{2}} (36 y^{2}) = 1 (36 y^{2})$

Step 5.3.2.1.4.2

Factor $4 y^{2}$ out of $36 y^{2}$ .

$4 y^{4} y' + \frac{- 9 y'}{4 y^{2}} (4 y^{2} (9)) = 1 (36 y^{2})$

Step 5.3.2.1.4.3

Cancel the common factor.

$4 y^{4} y' + \frac{- 9 y'}{4 y^{2}} (4 y^{2} \cdot 9) = 1 (36 y^{2})$

Step 5.3.2.1.4.4

Rewrite the expression.

$4 y^{4} y' - 9 y' \cdot 9 = 1 (36 y^{2})$

Step 5.3.2.1.5

Multiply $9$ by $- 9$ .

$4 y^{4} y' - 81 y' = 1 (36 y^{2})$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $36 y^{2}$ by $1$ .

$4 y^{4} y' - 81 y' = 36 y^{2}$

Step 5.4

Solve the equation.

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Step 5.4.1

Factor $y'$ out of $4 y^{4} y' - 81 y'$ .

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Step 5.4.1.1

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

$y' (4 y^{4}) - 81 y' = 36 y^{2}$

Step 5.4.1.2

Factor $y'$ out of $- 81 y'$ .

$y' (4 y^{4}) + y' \cdot - 81 = 36 y^{2}$

Step 5.4.1.3

Factor $y'$ out of $y' (4 y^{4}) + y' \cdot - 81$ .

$y' (4 y^{4} - 81) = 36 y^{2}$

Step 5.4.2

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

$y' ({(2 y^{2})}^{2} - 81) = 36 y^{2}$

Step 5.4.3

Rewrite $81$ as $9^{2}$ .

$y' ({(2 y^{2})}^{2} - 9^{2}) = 36 y^{2}$

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 $a = 2 y^{2}$ and $b = 9$ .

$y' ((2 y^{2} + 9) (2 y^{2} - 9)) = 36 y^{2}$

Step 5.4.4.2

Remove unnecessary parentheses.

$y' (2 y^{2} + 9) (2 y^{2} - 9) = 36 y^{2}$

Step 5.4.5

Divide each term in $y' (2 y^{2} + 9) (2 y^{2} - 9) = 36 y^{2}$ by $(2 y^{2} + 9) (2 y^{2} - 9)$ and simplify.

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Step 5.4.5.1

Divide each term in $y' (2 y^{2} + 9) (2 y^{2} - 9) = 36 y^{2}$ by $(2 y^{2} + 9) (2 y^{2} - 9)$ .

$\frac{y' (2 y^{2} + 9) (2 y^{2} - 9)}{(2 y^{2} + 9) (2 y^{2} - 9)} = \frac{36 y^{2}}{(2 y^{2} + 9) (2 y^{2} - 9)}$

Step 5.4.5.2

Simplify the left side.

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Step 5.4.5.2.1

Cancel the common factor of $2 y^{2} + 9$ .

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Step 5.4.5.2.1.1

Cancel the common factor.

$\frac{y' (2 y^{2} + 9) (2 y^{2} - 9)}{(2 y^{2} + 9) (2 y^{2} - 9)} = \frac{36 y^{2}}{(2 y^{2} + 9) (2 y^{2} - 9)}$

Step 5.4.5.2.1.2

Rewrite the expression.

$\frac{y' (2 y^{2} - 9)}{2 y^{2} - 9} = \frac{36 y^{2}}{(2 y^{2} + 9) (2 y^{2} - 9)}$

Step 5.4.5.2.2

Cancel the common factor of $2 y^{2} - 9$ .

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Step 5.4.5.2.2.1

Cancel the common factor.

$\frac{y' (2 y^{2} - 9)}{2 y^{2} - 9} = \frac{36 y^{2}}{(2 y^{2} + 9) (2 y^{2} - 9)}$

Step 5.4.5.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{36 y^{2}}{(2 y^{2} + 9) (2 y^{2} - 9)}$

Step 6

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

$\frac{d y}{d x} = \frac{36 y^{2}}{(2 y^{2} + 9) (2 y^{2} - 9)}$