Calculus Examples

$4 x^{2} + 9 y^{2} = 36$

Step 1

Solve the equation as $y$ in terms of $x$ .

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Step 1.1

Subtract $4 x^{2}$ from both sides of the equation.

$9 y^{2} = 36 - 4 x^{2}$

Step 1.2

Divide each term in $9 y^{2} = 36 - 4 x^{2}$ by $9$ and simplify.

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Step 1.2.1

Divide each term in $9 y^{2} = 36 - 4 x^{2}$ by $9$ .

$\frac{9 y^{2}}{9} = \frac{36}{9} + \frac{- 4 x^{2}}{9}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $9$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{9 y^{2}}{9} = \frac{36}{9} + \frac{- 4 x^{2}}{9}$

Step 1.2.2.1.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{36}{9} + \frac{- 4 x^{2}}{9}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide $36$ by $9$ .

$y^{2} = 4 + \frac{- 4 x^{2}}{9}$

Step 1.2.3.1.2

Move the negative in front of the fraction.

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

Step 1.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{4 - \frac{4 x^{2}}{9}}$

Step 1.4

Simplify $\pm \sqrt{4 - \frac{4 x^{2}}{9}}$ .

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Step 1.4.1

Write the expression using exponents.

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Step 1.4.1.1

Rewrite $4$ as $2^{2}$ .

$y = \pm \sqrt{2^{2} - \frac{4 x^{2}}{9}}$

Step 1.4.1.2

Rewrite $\frac{4 x^{2}}{9}$ as ${(\frac{2 x}{3})}^{2}$ .

$y = \pm \sqrt{2^{2} - {(\frac{2 x}{3})}^{2}}$

Step 1.4.2

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

$y = \pm \sqrt{(2 + \frac{2 x}{3}) (2 - \frac{2 x}{3})}$

Step 1.4.3

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \pm \sqrt{(2 \cdot \frac{3}{3} + \frac{2 x}{3}) (2 - \frac{2 x}{3})}$

Step 1.4.4

Combine $2$ and $\frac{3}{3}$ .

$y = \pm \sqrt{(\frac{2 \cdot 3}{3} + \frac{2 x}{3}) (2 - \frac{2 x}{3})}$

Step 1.4.5

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{2 \cdot 3 + 2 x}{3} (2 - \frac{2 x}{3})}$

Step 1.4.6

Factor $2$ out of $2 \cdot 3 + 2 x$ .

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Step 1.4.6.1

Factor $2$ out of $2 \cdot 3$ .

$y = \pm \sqrt{\frac{2 (3) + 2 x}{3} (2 - \frac{2 x}{3})}$

Step 1.4.6.2

Factor $2$ out of $2 x$ .

$y = \pm \sqrt{\frac{2 (3) + 2 (x)}{3} (2 - \frac{2 x}{3})}$

Step 1.4.6.3

Factor $2$ out of $2 (3) + 2 (x)$ .

$y = \pm \sqrt{\frac{2 (3 + x)}{3} (2 - \frac{2 x}{3})}$

Step 1.4.7

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \pm \sqrt{\frac{2 (3 + x)}{3} (2 \cdot \frac{3}{3} - \frac{2 x}{3})}$

Step 1.4.8

Combine $2$ and $\frac{3}{3}$ .

$y = \pm \sqrt{\frac{2 (3 + x)}{3} (\frac{2 \cdot 3}{3} - \frac{2 x}{3})}$

Step 1.4.9

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{2 (3 + x)}{3} \cdot \frac{2 \cdot 3 - 2 x}{3}}$

Step 1.4.10

Factor $2$ out of $2 \cdot 3 - 2 x$ .

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Step 1.4.10.1

Factor $2$ out of $2 \cdot 3$ .

$y = \pm \sqrt{\frac{2 (3 + x)}{3} \cdot \frac{2 (3) - 2 x}{3}}$

Step 1.4.10.2

Factor $2$ out of $- 2 x$ .

$y = \pm \sqrt{\frac{2 (3 + x)}{3} \cdot \frac{2 (3) + 2 (- x)}{3}}$

Step 1.4.10.3

Factor $2$ out of $2 (3) + 2 (- x)$ .

$y = \pm \sqrt{\frac{2 (3 + x)}{3} \cdot \frac{2 (3 - x)}{3}}$

Step 1.4.11

Multiply $\frac{2 (3 + x)}{3}$ by $\frac{2 (3 - x)}{3}$ .

$y = \pm \sqrt{\frac{2 (3 + x) (2 (3 - x))}{3 \cdot 3}}$

Step 1.4.12

Multiply.

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Step 1.4.12.1

Multiply $2$ by $2$ .

$y = \pm \sqrt{\frac{4 (3 + x) (3 - x)}{3 \cdot 3}}$

Step 1.4.12.2

Multiply $3$ by $3$ .

$y = \pm \sqrt{\frac{4 (3 + x) (3 - x)}{9}}$

Step 1.4.13

Rewrite $\frac{4 (3 + x) (3 - x)}{9}$ as ${(\frac{2}{3})}^{2} ((3 + x) (3 - x))$ .

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Step 1.4.13.1

Factor the perfect power $2^{2}$ out of $4 (3 + x) (3 - x)$ .

$y = \pm \sqrt{\frac{2^{2} ((3 + x) (3 - x))}{9}}$

Step 1.4.13.2

Factor the perfect power $3^{2}$ out of $9$ .

$y = \pm \sqrt{\frac{2^{2} ((3 + x) (3 - x))}{3^{2} \cdot 1}}$

Step 1.4.13.3

Rearrange the fraction $\frac{2^{2} ((3 + x) (3 - x))}{3^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{2}{3})}^{2} ((3 + x) (3 - x))}$

Step 1.4.14

Pull terms out from under the radical.

$y = \pm \frac{2}{3} \sqrt{(3 + x) (3 - x)}$

Step 1.4.15

Combine $\frac{2}{3}$ and $\sqrt{(3 + x) (3 - x)}$ .

$y = \pm \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

Step 1.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.5.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

Step 1.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

Step 1.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

$y = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

$y = \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

$y = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

$y = \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

$y = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

Step 2

Set each solution of $y$ as a function of $x$ .

$y = \frac{2 \sqrt{(3 + x) (3 - x)}}{3} \to f (x) = \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

$y = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3} \to f (x) = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$

Step 3

Because the $y$ variable in the equation $4 x^{2} + 9 y^{2} = 36$ has a degree greater than $1$ , use implicit differentiation to solve for the derivative $\frac{d y}{d x}$ .

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Step 3.1

Differentiate both sides of the equation.

$\frac{d}{d x} (4 x^{2} + 9 y^{2}) = \frac{d}{d x} (36)$

Step 3.2

Differentiate the left side of the equation.

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Step 3.2.1

By the Sum Rule, the derivative of $4 x^{2} + 9 y^{2}$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [9 y^{2}]$ .

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

Step 3.2.2

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

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Step 3.2.2.1

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

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

Step 3.2.2.2

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

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

Step 3.2.2.3

Multiply $2$ by $4$ .

$8 x + \frac{d}{d x} [9 y^{2}]$

Step 3.2.3

Evaluate $\frac{d}{d x} [9 y^{2}]$ .

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Step 3.2.3.1

Since $9$ is constant with respect to $x$ , the derivative of $9 y^{2}$ with respect to $x$ is $9 \frac{d}{d x} [y^{2}]$ .

$8 x + 9 \frac{d}{d x} [y^{2}]$

Step 3.2.3.2

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

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Step 3.2.3.2.1

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

$8 x + 9 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

Step 3.2.3.2.2

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

$8 x + 9 (2 u \frac{d}{d x} [y])$

Step 3.2.3.2.3

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

$8 x + 9 (2 y \frac{d}{d x} [y])$

Step 3.2.3.3

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

$8 x + 9 (2 y y')$

Step 3.2.3.4

Multiply $2$ by $9$ .

$8 x + 18 y y'$

Step 3.2.4

Reorder terms.

$18 y y' + 8 x$

Step 3.3

Since $36$ is constant with respect to $x$ , the derivative of $36$ with respect to $x$ is $0$ .

$0$

Step 3.4

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

$18 y y' + 8 x = 0$

Step 3.5

Solve for $y'$ .

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Step 3.5.1

Subtract $8 x$ from both sides of the equation.

$18 y y' = - 8 x$

Step 3.5.2

Divide each term in $18 y y' = - 8 x$ by $18 y$ and simplify.

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Step 3.5.2.1

Divide each term in $18 y y' = - 8 x$ by $18 y$ .

$\frac{18 y y'}{18 y} = \frac{- 8 x}{18 y}$

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $18$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

$\frac{18 y y'}{18 y} = \frac{- 8 x}{18 y}$

Step 3.5.2.2.1.2

Rewrite the expression.

$\frac{y y'}{y} = \frac{- 8 x}{18 y}$

Step 3.5.2.2.2

Cancel the common factor of $y$ .

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Step 3.5.2.2.2.1

Cancel the common factor.

$\frac{y y'}{y} = \frac{- 8 x}{18 y}$

Step 3.5.2.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 8 x}{18 y}$

Step 3.5.2.3

Simplify the right side.

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Step 3.5.2.3.1

Cancel the common factor of $- 8$ and $18$ .

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Step 3.5.2.3.1.1

Factor $2$ out of $- 8 x$ .

$y' = \frac{2 (- 4 x)}{18 y}$

Step 3.5.2.3.1.2

Cancel the common factors.

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Step 3.5.2.3.1.2.1

Factor $2$ out of $18 y$ .

$y' = \frac{2 (- 4 x)}{2 (9 y)}$

Step 3.5.2.3.1.2.2

Cancel the common factor.

$y' = \frac{2 (- 4 x)}{2 (9 y)}$

Step 3.5.2.3.1.2.3

Rewrite the expression.

$y' = \frac{- 4 x}{9 y}$

Step 3.5.2.3.2

Move the negative in front of the fraction.

$y' = - \frac{4 x}{9 y}$

Step 3.6

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

$\frac{d y}{d x} = - \frac{4 x}{9 y}$

Step 4

Set the derivative equal to $0$ then solve the equation $- \frac{4 x}{9 y} = 0$ .

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Step 4.1

Set the numerator equal to zero.

$4 x = 0$

Step 4.2

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 4.2.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $4$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $0$ by $4$ .

$x = 0$

Step 5

Solve the function $f (x) = \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$ at $x = 0$ .

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Step 5.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \frac{2 \sqrt{(3 + 0) (3 - (0))}}{3}$

Step 5.2

Simplify the result.

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Step 5.2.1

Remove parentheses.

$f (0) = \frac{2 \sqrt{(3 + 0) (3 - (0))}}{3}$

Step 5.2.2

Simplify the numerator.

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Step 5.2.2.1

Add $3$ and $0$ .

$f (0) = \frac{2 \sqrt{3 (3 - (0))}}{3}$

Step 5.2.2.2

Multiply $- 1$ by $0$ .

$f (0) = \frac{2 \sqrt{3 (3 + 0)}}{3}$

Step 5.2.2.3

Add $3$ and $0$ .

$f (0) = \frac{2 \sqrt{3 \cdot 3}}{3}$

Step 5.2.2.4

Multiply $3$ by $3$ .

$f (0) = \frac{2 \sqrt{9}}{3}$

Step 5.2.2.5

Rewrite $9$ as $3^{2}$ .

$f (0) = \frac{2 \sqrt{3^{2}}}{3}$

Step 5.2.2.6

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = \frac{2 \cdot 3}{3}$

Step 5.2.3

Simplify the expression.

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Step 5.2.3.1

Multiply $2$ by $3$ .

$f (0) = \frac{6}{3}$

Step 5.2.3.2

Divide $6$ by $3$ .

$f (0) = 2$

Step 5.2.4

The final answer is $2$ .

$2$

Step 6

Solve the function $f (x) = - \frac{2 \sqrt{(3 + x) (3 - x)}}{3}$ at $x = 0$ .

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Step 6.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = - \frac{2 \sqrt{(3 + 0) (3 - (0))}}{3}$

Step 6.2

Simplify the result.

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Step 6.2.1

Remove parentheses.

$f (0) = - \frac{2 \sqrt{(3 + 0) (3 - (0))}}{3}$

Step 6.2.2

Simplify the numerator.

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Step 6.2.2.1

Add $3$ and $0$ .

$f (0) = - \frac{2 \sqrt{3 (3 - (0))}}{3}$

Step 6.2.2.2

Multiply $- 1$ by $0$ .

$f (0) = - \frac{2 \sqrt{3 (3 + 0)}}{3}$

Step 6.2.2.3

Add $3$ and $0$ .

$f (0) = - \frac{2 \sqrt{3 \cdot 3}}{3}$

Step 6.2.2.4

Multiply $3$ by $3$ .

$f (0) = - \frac{2 \sqrt{9}}{3}$

Step 6.2.2.5

Rewrite $9$ as $3^{2}$ .

$f (0) = - \frac{2 \sqrt{3^{2}}}{3}$

Step 6.2.2.6

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = - \frac{2 \cdot 3}{3}$

Step 6.2.3

Simplify the expression.

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Step 6.2.3.1

Multiply $2$ by $3$ .

$f (0) = - \frac{6}{3}$

Step 6.2.3.2

Divide $6$ by $3$ .

$f (0) = - 1 \cdot 2$

Step 6.2.3.3

Multiply $- 1$ by $2$ .

$f (0) = - 2$

Step 6.2.4

The final answer is $- 2$ .

$- 2$

Step 7

The horizontal tangent lines are $y = 2, y = - 2$

$y = 2, y = - 2$

Step 8