Calculus Examples

$x^{3} + y^{3} - 9 x y = 0$

Step 1

Set each solution of $x^{3} + y^{3} - 9 x y$ as a function of $x$ .

$y = 0 \to f (x) = 0$

Step 2

Because the $y$ variable in the equation $x^{3} + y^{3} - 9 x y = 0$ has a degree greater than $1$ , use implicit differentiation to solve for the derivative $\frac{d y}{d x}$ .

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Step 2.1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{3} + y^{3} - 9 x y) = \frac{d}{d x} (0)$

Step 2.2

Differentiate the left side of the equation.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

By the Sum Rule, the derivative of $x^{3} + y^{3} - 9 x y$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [y^{3}] + \frac{d}{d x} [- 9 x y]$ .

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

Step 2.2.1.2

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

$3 x^{2} + \frac{d}{d x} [y^{3}] + \frac{d}{d x} [- 9 x y]$

Step 2.2.2

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

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

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Step 2.2.2.1.1

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

$3 x^{2} + \frac{d}{d u} [u^{3}] \frac{d}{d x} [y] + \frac{d}{d x} [- 9 x y]$

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 $n = 3$ .

$3 x^{2} + 3 u^{2} \frac{d}{d x} [y] + \frac{d}{d x} [- 9 x y]$

Step 2.2.2.1.3

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

$3 x^{2} + 3 y^{2} \frac{d}{d x} [y] + \frac{d}{d x} [- 9 x y]$

Step 2.2.2.2

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

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

Step 2.2.3

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

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Step 2.2.3.1

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

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

Step 2.2.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y$ .

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

Step 2.2.3.3

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

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

Step 2.2.3.4

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

$3 x^{2} + 3 y^{2} y' - 9 (x y' + y \cdot 1)$

Step 2.2.3.5

Multiply $y$ by $1$ .

$3 x^{2} + 3 y^{2} y' - 9 (x y' + y)$

Step 2.2.4

Simplify.

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Step 2.2.4.1

Apply the distributive property.

$3 x^{2} + 3 y^{2} y' - 9 (x y') - 9 y$

Step 2.2.4.2

Remove unnecessary parentheses.

$3 x^{2} + 3 y^{2} y' - 9 x y' - 9 y$

Step 2.3

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

$0$

Step 2.4

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

$3 x^{2} + 3 y^{2} y' - 9 x y' - 9 y = 0$

Step 2.5

Solve for $y'$ .

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Step 2.5.1

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

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Step 2.5.1.1

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

$3 y^{2} y' - 9 x y' - 9 y = - 3 x^{2}$

Step 2.5.1.2

Add $9 y$ to both sides of the equation.

$3 y^{2} y' - 9 x y' = - 3 x^{2} + 9 y$

Step 2.5.2

Factor $3 y'$ out of $3 y^{2} y' - 9 x y'$ .

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Step 2.5.2.1

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

$3 y' y^{2} - 9 x y' = - 3 x^{2} + 9 y$

Step 2.5.2.2

Factor $3 y'$ out of $- 9 x y'$ .

$3 y' y^{2} + 3 y' (- 3 x) = - 3 x^{2} + 9 y$

Step 2.5.2.3

Factor $3 y'$ out of $3 y' y^{2} + 3 y' (- 3 x)$ .

$3 y' (y^{2} - 3 x) = - 3 x^{2} + 9 y$

Step 2.5.3

Divide each term in $3 y' (y^{2} - 3 x) = - 3 x^{2} + 9 y$ by $3 (y^{2} - 3 x)$ and simplify.

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Step 2.5.3.1

Divide each term in $3 y' (y^{2} - 3 x) = - 3 x^{2} + 9 y$ by $3 (y^{2} - 3 x)$ .

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

Step 2.5.3.2

Simplify the left side.

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Step 2.5.3.2.1

Cancel the common factor of $3$ .

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Step 2.5.3.2.1.1

Cancel the common factor.

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

Step 2.5.3.2.1.2

Rewrite the expression.

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

Step 2.5.3.2.2

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

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Step 2.5.3.2.2.1

Cancel the common factor.

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

Step 2.5.3.2.2.2

Divide $y'$ by $1$ .

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

Step 2.5.3.3

Simplify the right side.

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Step 2.5.3.3.1

Simplify each term.

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Step 2.5.3.3.1.1

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

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Step 2.5.3.3.1.1.1

Factor $3$ out of $- 3 x^{2}$ .

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

Step 2.5.3.3.1.1.2

Cancel the common factors.

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Step 2.5.3.3.1.1.2.1

Cancel the common factor.

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

Step 2.5.3.3.1.1.2.2

Rewrite the expression.

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

Step 2.5.3.3.1.2

Move the negative in front of the fraction.

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

Step 2.5.3.3.1.3

Cancel the common factor of $9$ and $3$ .

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Step 2.5.3.3.1.3.1

Factor $3$ out of $9 y$ .

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

Step 2.5.3.3.1.3.2

Cancel the common factors.

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Step 2.5.3.3.1.3.2.1

Cancel the common factor.

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

Step 2.5.3.3.1.3.2.2

Rewrite the expression.

$y' = - \frac{x^{2}}{y^{2} - 3 x} + \frac{3 y}{y^{2} - 3 x}$

Step 2.5.3.3.2

Simplify terms.

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Step 2.5.3.3.2.1

Combine the numerators over the common denominator.

$y' = \frac{- x^{2} + 3 y}{y^{2} - 3 x}$

Step 2.5.3.3.2.2

Factor $- 1$ out of $- x^{2}$ .

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

Step 2.5.3.3.2.3

Factor $- 1$ out of $3 y$ .

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

Step 2.5.3.3.2.4

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

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

Step 2.5.3.3.2.5

Simplify the expression.

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Step 2.5.3.3.2.5.1

Rewrite $- (x^{2} - 3 y)$ as $- 1 (x^{2} - 3 y)$ .

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

Step 2.5.3.3.2.5.2

Move the negative in front of the fraction.

$y' = - \frac{x^{2} - 3 y}{y^{2} - 3 x}$

Step 2.6

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

$\frac{d y}{d x} = - \frac{x^{2} - 3 y}{y^{2} - 3 x}$

Step 3

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

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Step 3.1

Set the numerator equal to zero.

$x^{2} - 3 y = 0$

Step 3.2

Solve the equation for $x$ .

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Step 3.2.1

Add $3 y$ to both sides of the equation.

$x^{2} = 3 y$

Step 3.2.2

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

$x = \pm \sqrt{3 y}$

Step 3.2.3

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

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Step 3.2.3.1

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

$x = \sqrt{3 y}$

Step 3.2.3.2

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

$x = - \sqrt{3 y}$

Step 3.2.3.3

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

$x = \sqrt{3 y}$

$x = - \sqrt{3 y}$

$x = \sqrt{3 y}$

$x = - \sqrt{3 y}$

$x = \sqrt{3 y}$

$x = - \sqrt{3 y}$

$x = \sqrt{3 y}$

$x = - \sqrt{3 y}$

Step 4

Solve the function $f (x) = 0$ at $x = \sqrt{3 y}$ .

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Step 4.1

Replace the variable $x$ with $\sqrt{3 y}$ in the expression.

$f (\sqrt{3 y}) = 0$

Step 4.2

The final answer is $0$ .

$0$

Step 5

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

$y = 0, y = 0$

Step 6