Calculus Examples

$y^{3} - 27 y = x^{2} - 90$

Step 1

Set each solution of $y^{3} - 27 y$ as a function of $x$ .

$y = x^{2} - 90 \to f (x) = x^{2} - 90$

Step 2

Because the $y$ variable in the equation $y^{3} - 27 y = x^{2} - 90$ 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.

$\frac{d}{d x} (y^{3} - 27 y) = \frac{d}{d x} (x^{2} - 90)$

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

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

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 $y$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d x} [y] + \frac{d}{d x} [- 27 y]$

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 $n = 3$ .

$3 u^{2} \frac{d}{d x} [y] + \frac{d}{d x} [- 27 y]$

Step 2.2.2.1.3

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

$3 y^{2} \frac{d}{d x} [y] + \frac{d}{d x} [- 27 y]$

Step 2.2.2.2

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

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

Step 2.2.3

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

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Step 2.2.3.1

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

$3 y^{2} y' - 27 \frac{d}{d x} [y]$

Step 2.2.3.2

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

$3 y^{2} y' - 27 y'$

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 $x^{2} - 90$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 90]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 90]$

Step 2.3.2

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

$2 x + \frac{d}{d x} [- 90]$

Step 2.3.3

Since $- 90$ is constant with respect to $x$ , the derivative of $- 90$ with respect to $x$ is $0$ .

$2 x + 0$

Step 2.3.4

Add $2 x$ and $0$ .

$2 x$

Step 2.4

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

$3 y^{2} y' - 27 y' = 2 x$

Step 2.5

Solve for $y'$ .

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Step 2.5.1

Factor $3 y'$ out of $3 y^{2} y' - 27 y'$ .

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Step 2.5.1.1

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

$3 y' y^{2} - 27 y' = 2 x$

Step 2.5.1.2

Factor $3 y'$ out of $- 27 y'$ .

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

Step 2.5.1.3

Factor $3 y'$ out of $3 y' y^{2} + 3 y' \cdot - 9$ .

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

Step 2.5.2

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

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

Step 2.5.3

Factor.

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Step 2.5.3.1

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

$3 y' ((y + 3) (y - 3)) = 2 x$

Step 2.5.3.2

Remove unnecessary parentheses.

$3 y' (y + 3) (y - 3) = 2 x$

Step 2.5.4

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

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Step 2.5.4.1

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

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

Step 2.5.4.2

Simplify the left side.

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Step 2.5.4.2.1

Cancel the common factor of $3$ .

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Step 2.5.4.2.1.1

Cancel the common factor.

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

Step 2.5.4.2.1.2

Rewrite the expression.

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

Step 2.5.4.2.2

Cancel the common factor of $y + 3$ .

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Step 2.5.4.2.2.1

Cancel the common factor.

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

Step 2.5.4.2.2.2

Rewrite the expression.

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

Step 2.5.4.2.3

Cancel the common factor of $y - 3$ .

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Step 2.5.4.2.3.1

Cancel the common factor.

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

Step 2.5.4.2.3.2

Divide $y'$ by $1$ .

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

Step 2.6

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

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

Step 3

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

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Step 3.1

Set the numerator equal to zero.

$2 x = 0$

Step 3.2

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $0$ by $2$ .

$x = 0$

Step 4

Solve the function $f (x) = x^{2} - 90$ at $x = 0$ .

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Step 4.1

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

$f (0) = {(0)}^{2} - 90$

Step 4.2

Simplify the result.

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Step 4.2.1

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 - 90$

Step 4.2.2

Subtract $90$ from $0$ .

$f (0) = - 90$

Step 4.2.3

The final answer is $- 90$ .

$- 90$

Step 5

The horizontal tangent lines are $y = - 90$

$y = - 90$

Step 6