Calculus Examples

$y^{6} + x^{3} = y^{2} + 12 x$

Step 1

Set each solution of $y^{6} + x^{3}$ as a function of $x$ .

$y = y^{2} + 12 x \to f (x) = y^{2} + 12 x$

Step 2

Because the $y$ variable in the equation $y^{6} + x^{3} = y^{2} + 12 x$ has a degree greater than $1$ , use implicit differentiation to solve for the derivative $\frac{d y}{d x}$ .

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Step 2.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{6} + x^{3}) = \frac{d}{d x} (y^{2} + 12 x)$

Step 2.2

Differentiate the left side of the equation.

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Step 2.2.1

By the Sum Rule, the derivative of $y^{6} + x^{3}$ with respect to $x$ is $\frac{d}{d x} [y^{6}] + \frac{d}{d x} [x^{3}]$ .

$\frac{d}{d x} [y^{6}] + \frac{d}{d x} [x^{3}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [y^{6}]$ .

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Step 2.2.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{6}$ and $g (x) = y$ .

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Step 2.2.2.1.1

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

$\frac{d}{d u} [u^{6}] \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}]$

Step 2.2.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 6$ .

$6 u^{5} \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}]$

Step 2.2.2.1.3

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

$6 y^{5} \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}]$

Step 2.2.2.2

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

$6 y^{5} y' + \frac{d}{d x} [x^{3}]$

Step 2.2.3

Differentiate using the Power Rule.

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Step 2.2.3.1

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

$6 y^{5} y' + 3 x^{2}$

Step 2.2.3.2

Reorder terms.

$3 x^{2} + 6 y^{5} y'$

Step 2.3

Differentiate the right side of the equation.

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Step 2.3.1

By the Sum Rule, the derivative of $y^{2} + 12 x$ with respect to $x$ is $\frac{d}{d x} [y^{2}] + \frac{d}{d x} [12 x]$ .

$\frac{d}{d x} [y^{2}] + \frac{d}{d x} [12 x]$

Step 2.3.2

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

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Step 2.3.2.1

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 2.3.2.1.1

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

$\frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [12 x]$

Step 2.3.2.1.2

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

$2 u \frac{d}{d x} [y] + \frac{d}{d x} [12 x]$

Step 2.3.2.1.3

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

$2 y \frac{d}{d x} [y] + \frac{d}{d x} [12 x]$

Step 2.3.2.2

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

$2 y y' + \frac{d}{d x} [12 x]$

Step 2.3.3

Evaluate $\frac{d}{d x} [12 x]$ .

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Step 2.3.3.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x$ with respect to $x$ is $12 \frac{d}{d x} [x]$ .

$2 y y' + 12 \frac{d}{d x} [x]$

Step 2.3.3.2

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

$2 y y' + 12 \cdot 1$

Step 2.3.3.3

Multiply $12$ by $1$ .

$2 y y' + 12$

Step 2.4

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

$3 x^{2} + 6 y^{5} y' = 2 y y' + 12$

Step 2.5

Solve for $y'$ .

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Step 2.5.1

Subtract $2 y y'$ from both sides of the equation.

$3 x^{2} + 6 y^{5} y' - 2 y y' = 12$

Step 2.5.2

Subtract $3 x^{2}$ from both sides of the equation.

$6 y^{5} y' - 2 y y' = 12 - 3 x^{2}$

Step 2.5.3

Factor $2 y y'$ out of $6 y^{5} y' - 2 y y'$ .

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Step 2.5.3.1

Factor $2 y y'$ out of $6 y^{5} y'$ .

$2 y y' (3 y^{4}) - 2 y y' = 12 - 3 x^{2}$

Step 2.5.3.2

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

$2 y y' (3 y^{4}) + 2 y y' \cdot - 1 = 12 - 3 x^{2}$

Step 2.5.3.3

Factor $2 y y'$ out of $2 y y' (3 y^{4}) + 2 y y' \cdot - 1$ .

$2 y y' (3 y^{4} - 1) = 12 - 3 x^{2}$

Step 2.5.4

Divide each term in $2 y y' (3 y^{4} - 1) = 12 - 3 x^{2}$ by $2 y (3 y^{4} - 1)$ and simplify.

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Step 2.5.4.1

Divide each term in $2 y y' (3 y^{4} - 1) = 12 - 3 x^{2}$ by $2 y (3 y^{4} - 1)$ .

$\frac{2 y y' (3 y^{4} - 1)}{2 y (3 y^{4} - 1)} = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.2

Simplify the left side.

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Step 2.5.4.2.1

Cancel the common factor of $2$ .

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Step 2.5.4.2.1.1

Cancel the common factor.

$\frac{2 y y' (3 y^{4} - 1)}{2 y (3 y^{4} - 1)} = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.2.1.2

Rewrite the expression.

$\frac{y y' (3 y^{4} - 1)}{y (3 y^{4} - 1)} = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.2.2

Cancel the common factor of $y$ .

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Step 2.5.4.2.2.1

Cancel the common factor.

$\frac{y y' (3 y^{4} - 1)}{y (3 y^{4} - 1)} = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.2.2.2

Rewrite the expression.

$\frac{y' (3 y^{4} - 1)}{3 y^{4} - 1} = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.2.3

Cancel the common factor of $3 y^{4} - 1$ .

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Step 2.5.4.2.3.1

Cancel the common factor.

$\frac{y' (3 y^{4} - 1)}{3 y^{4} - 1} = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{12}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.3

Simplify the right side.

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Step 2.5.4.3.1

Simplify each term.

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Step 2.5.4.3.1.1

Cancel the common factor of $12$ and $2$ .

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Step 2.5.4.3.1.1.1

Factor $2$ out of $12$ .

$y' = \frac{2 \cdot 6}{2 y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.3.1.1.2

Cancel the common factors.

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Step 2.5.4.3.1.1.2.1

Factor $2$ out of $2 y (3 y^{4} - 1)$ .

$y' = \frac{2 \cdot 6}{2 (y (3 y^{4} - 1))} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.3.1.1.2.2

Cancel the common factor.

$y' = \frac{2 \cdot 6}{2 (y (3 y^{4} - 1))} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.3.1.1.2.3

Rewrite the expression.

$y' = \frac{6}{y (3 y^{4} - 1)} + \frac{- 3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.5.4.3.1.2

Move the negative in front of the fraction.

$y' = \frac{6}{y (3 y^{4} - 1)} - \frac{3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 2.6

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

$\frac{d y}{d x} = \frac{6}{y (3 y^{4} - 1)} - \frac{3 x^{2}}{2 y (3 y^{4} - 1)}$

Step 3

Set the derivative equal to $0$ then solve the equation $\frac{6}{y (3 y^{4} - 1)} - \frac{3 x^{2}}{2 y (3 y^{4} - 1)} = 0$ .

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Step 3.1

Subtract $\frac{6}{y (3 y^{4} - 1)}$ from both sides of the equation.

$- \frac{3 x^{2}}{2 y (3 y^{4} - 1)} = - \frac{6}{y (3 y^{4} - 1)}$

Step 3.2

Divide each term in $- \frac{3 x^{2}}{2 y (3 y^{4} - 1)} = - \frac{6}{y (3 y^{4} - 1)}$ by $- 1$ and simplify.

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Step 3.2.1

Divide each term in $- \frac{3 x^{2}}{2 y (3 y^{4} - 1)} = - \frac{6}{y (3 y^{4} - 1)}$ by $- 1$ .

$\frac{- \frac{3 x^{2}}{2 y (3 y^{4} - 1)}}{- 1} = \frac{- \frac{6}{y (3 y^{4} - 1)}}{- 1}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{3 x^{2}}{2 y (3 y^{4} - 1)}}{1} = \frac{- \frac{6}{y (3 y^{4} - 1)}}{- 1}$

Step 3.2.2.2

Divide $\frac{3 x^{2}}{2 y (3 y^{4} - 1)}$ by $1$ .

$\frac{3 x^{2}}{2 y (3 y^{4} - 1)} = \frac{- \frac{6}{y (3 y^{4} - 1)}}{- 1}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Dividing two negative values results in a positive value.

$\frac{3 x^{2}}{2 y (3 y^{4} - 1)} = \frac{\frac{6}{y (3 y^{4} - 1)}}{1}$

Step 3.2.3.2

Divide $\frac{6}{y (3 y^{4} - 1)}$ by $1$ .

$\frac{3 x^{2}}{2 y (3 y^{4} - 1)} = \frac{6}{y (3 y^{4} - 1)}$

Step 3.3

Multiply both sides by $2 y (3 y^{4} - 1)$ .

$\frac{3 x^{2}}{2 y (3 y^{4} - 1)} (2 y (3 y^{4} - 1)) = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the left side.

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Step 3.4.1.1

Simplify $\frac{3 x^{2}}{2 y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$ .

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Step 3.4.1.1.1

Rewrite using the commutative property of multiplication.

$2 \frac{3 x^{2}}{2 y (3 y^{4} - 1)} (y (3 y^{4} - 1)) = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4.1.1.2

Cancel the common factor of $2$ .

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Step 3.4.1.1.2.1

Factor $2$ out of $2 y (3 y^{4} - 1)$ .

$2 \frac{3 x^{2}}{2 (y (3 y^{4} - 1))} (y (3 y^{4} - 1)) = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4.1.1.2.2

Cancel the common factor.

$2 \frac{3 x^{2}}{2 (y (3 y^{4} - 1))} (y (3 y^{4} - 1)) = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4.1.1.2.3

Rewrite the expression.

$\frac{3 x^{2}}{y (3 y^{4} - 1)} (y (3 y^{4} - 1)) = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4.1.1.3

Cancel the common factor of $y (3 y^{4} - 1)$ .

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Step 3.4.1.1.3.1

Cancel the common factor.

$\frac{3 x^{2}}{y (3 y^{4} - 1)} (y (3 y^{4} - 1)) = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4.1.1.3.2

Rewrite the expression.

$3 x^{2} = \frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $\frac{6}{y (3 y^{4} - 1)} (2 y (3 y^{4} - 1))$ .

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Step 3.4.2.1.1

Rewrite using the commutative property of multiplication.

$3 x^{2} = 2 \frac{6}{y (3 y^{4} - 1)} (y (3 y^{4} - 1))$

Step 3.4.2.1.2

Multiply $2 \frac{6}{y (3 y^{4} - 1)}$ .

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Step 3.4.2.1.2.1

Combine $2$ and $\frac{6}{y (3 y^{4} - 1)}$ .

$3 x^{2} = \frac{2 \cdot 6}{y (3 y^{4} - 1)} (y (3 y^{4} - 1))$

Step 3.4.2.1.2.2

Multiply $2$ by $6$ .

$3 x^{2} = \frac{12}{y (3 y^{4} - 1)} (y (3 y^{4} - 1))$

Step 3.4.2.1.3

Cancel the common factor of $y (3 y^{4} - 1)$ .

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Step 3.4.2.1.3.1

Cancel the common factor.

$3 x^{2} = \frac{12}{y (3 y^{4} - 1)} (y (3 y^{4} - 1))$

Step 3.4.2.1.3.2

Rewrite the expression.

$3 x^{2} = 12$

Step 3.5

Solve for $x$ .

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Step 3.5.1

Divide each term in $3 x^{2} = 12$ by $3$ and simplify.

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Step 3.5.1.1

Divide each term in $3 x^{2} = 12$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Step 3.5.1.2

Simplify the left side.

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Step 3.5.1.2.1

Cancel the common factor of $3$ .

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Step 3.5.1.2.1.1

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Step 3.5.1.2.1.2

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

$x^{2} = \frac{12}{3}$

Step 3.5.1.3

Simplify the right side.

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Step 3.5.1.3.1

Divide $12$ by $3$ .

$x^{2} = 4$

Step 3.5.2

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

$x = \pm \sqrt{4}$

Step 3.5.3

Simplify $\pm \sqrt{4}$ .

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Step 3.5.3.1

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

$x = \pm \sqrt{2^{2}}$

Step 3.5.3.2

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

$x = \pm 2$

Step 3.5.4

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

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Step 3.5.4.1

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

$x = 2$

Step 3.5.4.2

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

$x = - 2$

Step 3.5.4.3

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

$x = 2, - 2$

Step 4

Solve the function $f (x) = y^{2} + 12 x$ at $x = 2$ .

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Step 4.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = y^{2} + 12 (2)$

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $12$ by $2$ .

$f (2) = y^{2} + 24$

Step 4.2.2

The final answer is $y^{2} + 24$ .

$y^{2} + 24$

Step 5

Solve the function $f (x) = y^{2} + 12 x$ at $x = - 2$ .

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Step 5.1

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = y^{2} + 12 (- 2)$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $12$ by $- 2$ .

$f (- 2) = y^{2} - 24$

Step 5.2.2

The final answer is $y^{2} - 24$ .

$y^{2} - 24$

Step 6

The horizontal tangent lines are $y = y^{2} + 24, y = y^{2} - 24$

$y = y^{2} + 24, y = y^{2} - 24$

Step 7