Calculus Examples

$4 x^{2} + y^{2} - 8 x + 4 y + 4 = 0$

Step 1

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

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Step 1.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 1.2

Substitute the values $a = 1$ , $b = 4$ , and $c = 4 x^{2} - 8 x + 4$ into the quadratic formula and solve for $y$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot (4 x^{2} - 8 x + 4))}}{2 \cdot 1}$

Step 1.3

Simplify.

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Step 1.3.1

Simplify the numerator.

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Step 1.3.1.1

Rewrite $4 \cdot 1 \cdot (4 x^{2} - 8 x + 4)$ as ${(2 \cdot 1 \cdot (2 x - 2))}^{2}$ .

$y = \frac{- 4 \pm \sqrt{4^{2} - {(2 \cdot 1 \cdot (2 x - 2))}^{2}}}{2 \cdot 1}$

Step 1.3.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = 2 \cdot 1 \cdot (2 x - 2)$ .

$y = \frac{- 4 \pm \sqrt{(4 + 2 \cdot 1 \cdot (2 x - 2)) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3

Simplify.

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Step 1.3.1.3.1

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm \sqrt{(4 + 2 \cdot (2 x - 2)) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.2

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{(4 + 2 (2 x) + 2 \cdot - 2) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.3

Multiply $2$ by $2$ .

$y = \frac{- 4 \pm \sqrt{(4 + 4 x + 2 \cdot - 2) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.4

Multiply $2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{(4 + 4 x - 4) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.5

Subtract $4$ from $4$ .

$y = \frac{- 4 \pm \sqrt{(4 x + 0) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.6

Add $4 x$ and $0$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.7

Combine exponents.

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Step 1.3.1.3.7.1

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - (2 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.3.1.3.7.2

Multiply $2$ by $- 1$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 2 (2 x - 2))}}{2 \cdot 1}$

Step 1.3.1.4

Simplify each term.

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Step 1.3.1.4.1

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{4 x (4 - 2 (2 x) - 2 \cdot - 2)}}{2 \cdot 1}$

Step 1.3.1.4.2

Multiply $2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 4 x - 2 \cdot - 2)}}{2 \cdot 1}$

Step 1.3.1.4.3

Multiply $- 2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 4 x + 4)}}{2 \cdot 1}$

Step 1.3.1.5

Add $4$ and $4$ .

$y = \frac{- 4 \pm \sqrt{4 x (- 4 x + 8)}}{2 \cdot 1}$

Step 1.3.1.6

Factor $4$ out of $- 4 x + 8$ .

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Step 1.3.1.6.1

Factor $4$ out of $- 4 x$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x) + 8)}}{2 \cdot 1}$

Step 1.3.1.6.2

Factor $4$ out of $8$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x) + 4 (2))}}{2 \cdot 1}$

Step 1.3.1.6.3

Factor $4$ out of $4 (- x) + 4 (2)$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x + 2))}}{2 \cdot 1}$

$y = \frac{- 4 \pm \sqrt{4 x \cdot (4 (- x + 2))}}{2 \cdot 1}$

Step 1.3.1.7

Multiply $4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 x (- x + 2)}}{2 \cdot 1}$

Step 1.3.1.8

Rewrite $16 x (- x + 2)$ as ${(2^{2})}^{2} (x (- x + 2))$ .

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Step 1.3.1.8.1

Rewrite $16$ as $4^{2}$ .

$y = \frac{- 4 \pm \sqrt{4^{2} x (- x + 2)}}{2 \cdot 1}$

Step 1.3.1.8.2

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

$y = \frac{- 4 \pm \sqrt{{(2^{2})}^{2} x (- x + 2)}}{2 \cdot 1}$

Step 1.3.1.8.3

Add parentheses.

$y = \frac{- 4 \pm \sqrt{{(2^{2})}^{2} (x (- x + 2))}}{2 \cdot 1}$

Step 1.3.1.9

Pull terms out from under the radical.

$y = \frac{- 4 \pm 2^{2} \sqrt{x (- x + 2)}}{2 \cdot 1}$

Step 1.3.1.10

Raise $2$ to the power of $2$ .

$y = \frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2 \cdot 1}$

Step 1.3.2

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2}$

Step 1.3.3

Simplify $\frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2}$ .

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

Step 1.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.4.1

Simplify the numerator.

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Step 1.4.1.1

Rewrite $4 \cdot 1 \cdot (4 x^{2} - 8 x + 4)$ as ${(2 \cdot 1 \cdot (2 x - 2))}^{2}$ .

$y = \frac{- 4 \pm \sqrt{4^{2} - {(2 \cdot 1 \cdot (2 x - 2))}^{2}}}{2 \cdot 1}$

Step 1.4.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = 2 \cdot 1 \cdot (2 x - 2)$ .

$y = \frac{- 4 \pm \sqrt{(4 + 2 \cdot 1 \cdot (2 x - 2)) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3

Simplify.

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Step 1.4.1.3.1

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm \sqrt{(4 + 2 \cdot (2 x - 2)) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.2

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{(4 + 2 (2 x) + 2 \cdot - 2) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.3

Multiply $2$ by $2$ .

$y = \frac{- 4 \pm \sqrt{(4 + 4 x + 2 \cdot - 2) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.4

Multiply $2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{(4 + 4 x - 4) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.5

Subtract $4$ from $4$ .

$y = \frac{- 4 \pm \sqrt{(4 x + 0) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.6

Add $4 x$ and $0$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.7

Combine exponents.

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Step 1.4.1.3.7.1

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - (2 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.4.1.3.7.2

Multiply $2$ by $- 1$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 2 (2 x - 2))}}{2 \cdot 1}$

Step 1.4.1.4

Simplify each term.

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Step 1.4.1.4.1

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{4 x (4 - 2 (2 x) - 2 \cdot - 2)}}{2 \cdot 1}$

Step 1.4.1.4.2

Multiply $2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 4 x - 2 \cdot - 2)}}{2 \cdot 1}$

Step 1.4.1.4.3

Multiply $- 2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 4 x + 4)}}{2 \cdot 1}$

Step 1.4.1.5

Add $4$ and $4$ .

$y = \frac{- 4 \pm \sqrt{4 x (- 4 x + 8)}}{2 \cdot 1}$

Step 1.4.1.6

Factor $4$ out of $- 4 x + 8$ .

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Step 1.4.1.6.1

Factor $4$ out of $- 4 x$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x) + 8)}}{2 \cdot 1}$

Step 1.4.1.6.2

Factor $4$ out of $8$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x) + 4 (2))}}{2 \cdot 1}$

Step 1.4.1.6.3

Factor $4$ out of $4 (- x) + 4 (2)$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x + 2))}}{2 \cdot 1}$

$y = \frac{- 4 \pm \sqrt{4 x \cdot (4 (- x + 2))}}{2 \cdot 1}$

Step 1.4.1.7

Multiply $4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 x (- x + 2)}}{2 \cdot 1}$

Step 1.4.1.8

Rewrite $16 x (- x + 2)$ as ${(2^{2})}^{2} (x (- x + 2))$ .

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Step 1.4.1.8.1

Rewrite $16$ as $4^{2}$ .

$y = \frac{- 4 \pm \sqrt{4^{2} x (- x + 2)}}{2 \cdot 1}$

Step 1.4.1.8.2

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

$y = \frac{- 4 \pm \sqrt{{(2^{2})}^{2} x (- x + 2)}}{2 \cdot 1}$

Step 1.4.1.8.3

Add parentheses.

$y = \frac{- 4 \pm \sqrt{{(2^{2})}^{2} (x (- x + 2))}}{2 \cdot 1}$

Step 1.4.1.9

Pull terms out from under the radical.

$y = \frac{- 4 \pm 2^{2} \sqrt{x (- x + 2)}}{2 \cdot 1}$

Step 1.4.1.10

Raise $2$ to the power of $2$ .

$y = \frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2 \cdot 1}$

Step 1.4.2

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2}$

Step 1.4.3

Simplify $\frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2}$ .

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

Step 1.4.4

Change the $\pm$ to $+$ .

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

Step 1.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.5.1

Simplify the numerator.

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Step 1.5.1.1

Rewrite $4 \cdot 1 \cdot (4 x^{2} - 8 x + 4)$ as ${(2 \cdot 1 \cdot (2 x - 2))}^{2}$ .

$y = \frac{- 4 \pm \sqrt{4^{2} - {(2 \cdot 1 \cdot (2 x - 2))}^{2}}}{2 \cdot 1}$

Step 1.5.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = 2 \cdot 1 \cdot (2 x - 2)$ .

$y = \frac{- 4 \pm \sqrt{(4 + 2 \cdot 1 \cdot (2 x - 2)) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3

Simplify.

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Step 1.5.1.3.1

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm \sqrt{(4 + 2 \cdot (2 x - 2)) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.2

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{(4 + 2 (2 x) + 2 \cdot - 2) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.3

Multiply $2$ by $2$ .

$y = \frac{- 4 \pm \sqrt{(4 + 4 x + 2 \cdot - 2) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.4

Multiply $2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{(4 + 4 x - 4) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.5

Subtract $4$ from $4$ .

$y = \frac{- 4 \pm \sqrt{(4 x + 0) (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.6

Add $4 x$ and $0$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - (2 \cdot 1 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.7

Combine exponents.

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Step 1.5.1.3.7.1

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - (2 \cdot (2 x - 2)))}}{2 \cdot 1}$

Step 1.5.1.3.7.2

Multiply $2$ by $- 1$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 2 (2 x - 2))}}{2 \cdot 1}$

Step 1.5.1.4

Simplify each term.

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Step 1.5.1.4.1

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{4 x (4 - 2 (2 x) - 2 \cdot - 2)}}{2 \cdot 1}$

Step 1.5.1.4.2

Multiply $2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 4 x - 2 \cdot - 2)}}{2 \cdot 1}$

Step 1.5.1.4.3

Multiply $- 2$ by $- 2$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 - 4 x + 4)}}{2 \cdot 1}$

Step 1.5.1.5

Add $4$ and $4$ .

$y = \frac{- 4 \pm \sqrt{4 x (- 4 x + 8)}}{2 \cdot 1}$

Step 1.5.1.6

Factor $4$ out of $- 4 x + 8$ .

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Step 1.5.1.6.1

Factor $4$ out of $- 4 x$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x) + 8)}}{2 \cdot 1}$

Step 1.5.1.6.2

Factor $4$ out of $8$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x) + 4 (2))}}{2 \cdot 1}$

Step 1.5.1.6.3

Factor $4$ out of $4 (- x) + 4 (2)$ .

$y = \frac{- 4 \pm \sqrt{4 x (4 (- x + 2))}}{2 \cdot 1}$

$y = \frac{- 4 \pm \sqrt{4 x \cdot (4 (- x + 2))}}{2 \cdot 1}$

Step 1.5.1.7

Multiply $4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 x (- x + 2)}}{2 \cdot 1}$

Step 1.5.1.8

Rewrite $16 x (- x + 2)$ as ${(2^{2})}^{2} (x (- x + 2))$ .

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Step 1.5.1.8.1

Rewrite $16$ as $4^{2}$ .

$y = \frac{- 4 \pm \sqrt{4^{2} x (- x + 2)}}{2 \cdot 1}$

Step 1.5.1.8.2

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

$y = \frac{- 4 \pm \sqrt{{(2^{2})}^{2} x (- x + 2)}}{2 \cdot 1}$

Step 1.5.1.8.3

Add parentheses.

$y = \frac{- 4 \pm \sqrt{{(2^{2})}^{2} (x (- x + 2))}}{2 \cdot 1}$

Step 1.5.1.9

Pull terms out from under the radical.

$y = \frac{- 4 \pm 2^{2} \sqrt{x (- x + 2)}}{2 \cdot 1}$

Step 1.5.1.10

Raise $2$ to the power of $2$ .

$y = \frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2 \cdot 1}$

Step 1.5.2

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2}$

Step 1.5.3

Simplify $\frac{- 4 \pm 4 \sqrt{x (- x + 2)}}{2}$ .

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

Step 1.5.4

Change the $\pm$ to $-$ .

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

Step 1.6

The final answer is the combination of both solutions.

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

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

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

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

Step 2

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

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

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

Step 3

Because the $y$ variable in the equation $4 x^{2} + y^{2} - 8 x + 4 y + 4 = 0$ 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.

$\frac{d}{d x} (4 x^{2} + y^{2} - 8 x + 4 y + 4) = \frac{d}{d x} (0)$

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 $4 x^{2} + y^{2} - 8 x + 4 y + 4$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [4]$ .

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

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

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

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 $n = 2$ .

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

Step 3.2.2.3

Multiply $2$ by $4$ .

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

Step 3.2.3

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

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Step 3.2.3.1

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 3.2.3.1.1

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

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

Step 3.2.3.1.2

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

$8 x + 2 u \frac{d}{d x} [y] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [4]$

Step 3.2.3.1.3

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

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

Step 3.2.3.2

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

$8 x + 2 y y' + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [4]$

Step 3.2.4

Evaluate $\frac{d}{d x} [- 8 x]$ .

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Step 3.2.4.1

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

$8 x + 2 y y' - 8 \frac{d}{d x} [x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [4]$

Step 3.2.4.2

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

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

Step 3.2.4.3

Multiply $- 8$ by $1$ .

$8 x + 2 y y' - 8 + \frac{d}{d x} [4 y] + \frac{d}{d x} [4]$

Step 3.2.5

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

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Step 3.2.5.1

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

$8 x + 2 y y' - 8 + 4 \frac{d}{d x} [y] + \frac{d}{d x} [4]$

Step 3.2.5.2

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

$8 x + 2 y y' - 8 + 4 y' + \frac{d}{d x} [4]$

Step 3.2.6

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$8 x + 2 y y' - 8 + 4 y' + 0$

Step 3.2.7

Simplify.

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Step 3.2.7.1

Add $8 x + 2 y y' - 8 + 4 y'$ and $0$ .

$8 x + 2 y y' - 8 + 4 y'$

Step 3.2.7.2

Reorder terms.

$2 y y' + 8 x + 4 y' - 8$

Step 3.3

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

$0$

Step 3.4

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

$2 y y' + 8 x + 4 y' - 8 = 0$

Step 3.5

Solve for $y'$ .

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Step 3.5.1

Move all terms not containing $y'$ to the right side of the equation.

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Step 3.5.1.1

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

$2 y y' + 4 y' - 8 = - 8 x$

Step 3.5.1.2

Add $8$ to both sides of the equation.

$2 y y' + 4 y' = - 8 x + 8$

Step 3.5.2

Factor $2 y'$ out of $2 y y' + 4 y'$ .

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Step 3.5.2.1

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

$2 y' y + 4 y' = - 8 x + 8$

Step 3.5.2.2

Factor $2 y'$ out of $4 y'$ .

$2 y' y + 2 y' \cdot 2 = - 8 x + 8$

Step 3.5.2.3

Factor $2 y'$ out of $2 y' y + 2 y' \cdot 2$ .

$2 y' (y + 2) = - 8 x + 8$

Step 3.5.3

Divide each term in $2 y' (y + 2) = - 8 x + 8$ by $2 (y + 2)$ and simplify.

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Step 3.5.3.1

Divide each term in $2 y' (y + 2) = - 8 x + 8$ by $2 (y + 2)$ .

$\frac{2 y' (y + 2)}{2 (y + 2)} = \frac{- 8 x}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $2$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{2 y' (y + 2)}{2 (y + 2)} = \frac{- 8 x}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.2.1.2

Rewrite the expression.

$\frac{y' (y + 2)}{y + 2} = \frac{- 8 x}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.2.2

Cancel the common factor of $y + 2$ .

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Step 3.5.3.2.2.1

Cancel the common factor.

$\frac{y' (y + 2)}{y + 2} = \frac{- 8 x}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 8 x}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Simplify each term.

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Step 3.5.3.3.1.1

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

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Step 3.5.3.3.1.1.1

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

$y' = \frac{2 (- 4 x)}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.3.1.1.2

Cancel the common factors.

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Step 3.5.3.3.1.1.2.1

Cancel the common factor.

$y' = \frac{2 (- 4 x)}{2 (y + 2)} + \frac{8}{2 (y + 2)}$

Step 3.5.3.3.1.1.2.2

Rewrite the expression.

$y' = \frac{- 4 x}{y + 2} + \frac{8}{2 (y + 2)}$

Step 3.5.3.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{(4) x}{y + 2} + \frac{8}{2 (y + 2)}$

Step 3.5.3.3.1.3

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

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Step 3.5.3.3.1.3.1

Factor $2$ out of $8$ .

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

Step 3.5.3.3.1.3.2

Cancel the common factors.

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Step 3.5.3.3.1.3.2.1

Cancel the common factor.

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

Step 3.5.3.3.1.3.2.2

Rewrite the expression.

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

Step 3.5.3.3.2

Simplify terms.

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Step 3.5.3.3.2.1

Combine the numerators over the common denominator.

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

Step 3.5.3.3.2.2

Factor $4$ out of $- 4 x + 4$ .

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Step 3.5.3.3.2.2.1

Factor $4$ out of $- 4 x$ .

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

Step 3.5.3.3.2.2.2

Factor $4$ out of $4$ .

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

Step 3.5.3.3.2.2.3

Factor $4$ out of $4 (- x) + 4 (1)$ .

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

Step 3.5.3.3.2.3

Factor $- 1$ out of $- x$ .

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

Step 3.5.3.3.2.4

Rewrite $1$ as $- 1 (- 1)$ .

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

Step 3.5.3.3.2.5

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

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

Step 3.5.3.3.2.6

Simplify the expression.

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Step 3.5.3.3.2.6.1

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

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

Step 3.5.3.3.2.6.2

Move the negative in front of the fraction.

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

Step 3.6

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

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

Step 4

Set the derivative equal to $0$ then solve the equation $- \frac{4 (x - 1)}{y + 2} = 0$ .

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Step 4.1

Set the numerator equal to zero.

$4 (x - 1) = 0$

Step 4.2

Solve the equation for $x$ .

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Step 4.2.1

Divide each term in $4 (x - 1) = 0$ by $4$ and simplify.

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Step 4.2.1.1

Divide each term in $4 (x - 1) = 0$ by $4$ .

$\frac{4 (x - 1)}{4} = \frac{0}{4}$

Step 4.2.1.2

Simplify the left side.

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Step 4.2.1.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.2.1.1

Cancel the common factor.

$\frac{4 (x - 1)}{4} = \frac{0}{4}$

Step 4.2.1.2.1.2

Divide $x - 1$ by $1$ .

$x - 1 = \frac{0}{4}$

Step 4.2.1.3

Simplify the right side.

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Step 4.2.1.3.1

Divide $0$ by $4$ .

$x - 1 = 0$

Step 4.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Solve the function $f (x) = - 2 + 2 \sqrt{x (- x + 2)}$ at $x = 1$ .

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Step 5.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = - 2 + 2 \sqrt{(1) (- (1) + 2)}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $- (1) + 2$ by $1$ .

$f (1) = - 2 + 2 \sqrt{- (1) + 2}$

Step 5.2.1.2

Multiply $- 1$ by $1$ .

$f (1) = - 2 + 2 \sqrt{- 1 + 2}$

Step 5.2.1.3

Add $- 1$ and $2$ .

$f (1) = - 2 + 2 \sqrt{1}$

Step 5.2.1.4

Any root of $1$ is $1$ .

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

Step 5.2.1.5

Multiply $2$ by $1$ .

$f (1) = - 2 + 2$

Step 5.2.2

Add $- 2$ and $2$ .

$f (1) = 0$

Step 5.2.3

The final answer is $0$ .

$0$

Step 6

Solve the function $f (x) = - 2 - 2 \sqrt{x (- x + 2)}$ at $x = 1$ .

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = - 2 - 2 \sqrt{(1) (- (1) + 2)}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $- (1) + 2$ by $1$ .

$f (1) = - 2 - 2 \sqrt{- (1) + 2}$

Step 6.2.1.2

Multiply $- 1$ by $1$ .

$f (1) = - 2 - 2 \sqrt{- 1 + 2}$

Step 6.2.1.3

Add $- 1$ and $2$ .

$f (1) = - 2 - 2 \sqrt{1}$

Step 6.2.1.4

Any root of $1$ is $1$ .

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

Step 6.2.1.5

Multiply $- 2$ by $1$ .

$f (1) = - 2 - 2$

Step 6.2.2

Subtract $2$ from $- 2$ .

$f (1) = - 4$

Step 6.2.3

The final answer is $- 4$ .

$- 4$

Step 7

The horizontal tangent lines are $y = 0, y = - 4$

$y = 0, y = - 4$

Step 8