Calculus Examples

$y^{2} - x y - 12 = 0$

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 $a = 1$ , $b = - x$ , and $c = - 12$ into the quadratic formula and solve for $y$ .

$\frac{x \pm \sqrt{{(- x)}^{2} - 4 \cdot (1 \cdot - 12)}}{2 \cdot 1}$

Step 1.3

Simplify.

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Step 1.3.1

Simplify the numerator.

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Step 1.3.1.1

Apply the product rule to $- x$ .

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

Step 1.3.1.2

Raise $- 1$ to the power of $2$ .

$y = \frac{x \pm \sqrt{1 x^{2} - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Step 1.3.1.3

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

$y = \frac{x \pm \sqrt{x^{2} - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Step 1.3.1.4

Multiply $- 4 \cdot 1 \cdot - 12$ .

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Step 1.3.1.4.1

Multiply $- 4$ by $1$ .

$y = \frac{x \pm \sqrt{x^{2} - 4 \cdot - 12}}{2 \cdot 1}$

Step 1.3.1.4.2

Multiply $- 4$ by $- 12$ .

$y = \frac{x \pm \sqrt{x^{2} + 48}}{2 \cdot 1}$

Step 1.3.2

Multiply $2$ by $1$ .

$y = \frac{x \pm \sqrt{x^{2} + 48}}{2}$

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

Apply the product rule to $- x$ .

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

Step 1.4.1.2

Raise $- 1$ to the power of $2$ .

$y = \frac{x \pm \sqrt{1 x^{2} - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Step 1.4.1.3

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

$y = \frac{x \pm \sqrt{x^{2} - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Step 1.4.1.4

Multiply $- 4 \cdot 1 \cdot - 12$ .

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Step 1.4.1.4.1

Multiply $- 4$ by $1$ .

$y = \frac{x \pm \sqrt{x^{2} - 4 \cdot - 12}}{2 \cdot 1}$

Step 1.4.1.4.2

Multiply $- 4$ by $- 12$ .

$y = \frac{x \pm \sqrt{x^{2} + 48}}{2 \cdot 1}$

Step 1.4.2

Multiply $2$ by $1$ .

$y = \frac{x \pm \sqrt{x^{2} + 48}}{2}$

Step 1.4.3

Change the $\pm$ to $+$ .

$y = \frac{x + \sqrt{x^{2} + 48}}{2}$

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

Apply the product rule to $- x$ .

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

Step 1.5.1.2

Raise $- 1$ to the power of $2$ .

$y = \frac{x \pm \sqrt{1 x^{2} - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Step 1.5.1.3

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

$y = \frac{x \pm \sqrt{x^{2} - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Step 1.5.1.4

Multiply $- 4 \cdot 1 \cdot - 12$ .

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Step 1.5.1.4.1

Multiply $- 4$ by $1$ .

$y = \frac{x \pm \sqrt{x^{2} - 4 \cdot - 12}}{2 \cdot 1}$

Step 1.5.1.4.2

Multiply $- 4$ by $- 12$ .

$y = \frac{x \pm \sqrt{x^{2} + 48}}{2 \cdot 1}$

Step 1.5.2

Multiply $2$ by $1$ .

$y = \frac{x \pm \sqrt{x^{2} + 48}}{2}$

Step 1.5.3

Change the $\pm$ to $-$ .

$y = \frac{x - \sqrt{x^{2} + 48}}{2}$

Step 1.6

The final answer is the combination of both solutions.

$y = \frac{x + \sqrt{x^{2} + 48}}{2}$

$y = \frac{x - \sqrt{x^{2} + 48}}{2}$

$y = \frac{x + \sqrt{x^{2} + 48}}{2}$

$y = \frac{x - \sqrt{x^{2} + 48}}{2}$

Step 2

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

$y = \frac{x + \sqrt{x^{2} + 48}}{2} \to f (x) = \frac{x + \sqrt{x^{2} + 48}}{2}$

$y = \frac{x - \sqrt{x^{2} + 48}}{2} \to f (x) = \frac{x - \sqrt{x^{2} + 48}}{2}$

Step 3

Because the $y$ variable in the equation $y^{2} - x y - 12 = 0$ 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.

$\frac{d}{d x} (y^{2} - x y - 12) = \frac{d}{d x} (0)$

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 $y^{2} - x y - 12$ with respect to $x$ is $\frac{d}{d x} [y^{2}] + \frac{d}{d x} [- x y] + \frac{d}{d x} [- 12]$ .

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

Step 3.2.2

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

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Step 3.2.2.1

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

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

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

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

Step 3.2.2.1.3

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

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

Step 3.2.2.2

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

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

Step 3.2.3

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

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Step 3.2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x y$ with respect to $x$ is $- \frac{d}{d x} [x y]$ .

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

Step 3.2.3.2

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 $g (x) = y$ .

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

Step 3.2.3.3

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

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

Step 3.2.3.4

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

$2 y y' - (x y' + y \cdot 1) + \frac{d}{d x} [- 12]$

Step 3.2.3.5

Multiply $y$ by $1$ .

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

Step 3.2.4

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$2 y y' - (x y' + y) + 0$

Step 3.2.5

Simplify.

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Step 3.2.5.1

Apply the distributive property.

$2 y y' - (x y') - y + 0$

Step 3.2.5.2

Add $2 y y' - x y' - y$ and $0$ .

$2 y y' - x y' - y$

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.

$2 y y' - x y' - y = 0$

Step 3.5

Solve for $y'$ .

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Step 3.5.1

Add $y$ to both sides of the equation.

$2 y y' - x y' = y$

Step 3.5.2

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

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Step 3.5.2.1

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

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

Step 3.5.2.2

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

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

Step 3.5.2.3

Factor $y'$ out of $y' (2 y) + y' (- x)$ .

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

Step 3.5.3

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

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Step 3.5.3.1

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

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

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $2 y - x$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

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

Step 3.5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 3.6

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

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

Step 4

The roots of the derivative $\frac{y}{2 y - x}$ cannot be found.

No horizontal tangent lines

Step 5