Calculus Examples

$4 x + 5 y^{2} - y = 3$

Step 1

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

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Step 1.1

Subtract $3$ from both sides of the equation.

$4 x + 5 y^{2} - y - 3 = 0$

Step 1.2

Use the quadratic formula to find the solutions.

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

Step 1.3

Substitute the values $a = 5$ , $b = - 1$ , and $c = 4 x - 3$ into the quadratic formula and solve for $y$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (5 \cdot (4 x - 3))}}{2 \cdot 5}$

Step 1.4

Simplify.

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Step 1.4.1

Simplify the numerator.

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Step 1.4.1.1

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

$y = \frac{1 \pm \sqrt{1 - 4 \cdot 5 \cdot (4 x - 3)}}{2 \cdot 5}$

Step 1.4.1.2

Multiply $- 4$ by $5$ .

$y = \frac{1 \pm \sqrt{1 - 20 \cdot (4 x - 3)}}{2 \cdot 5}$

Step 1.4.1.3

Apply the distributive property.

$y = \frac{1 \pm \sqrt{1 - 20 (4 x) - 20 \cdot - 3}}{2 \cdot 5}$

Step 1.4.1.4

Multiply $4$ by $- 20$ .

$y = \frac{1 \pm \sqrt{1 - 80 x - 20 \cdot - 3}}{2 \cdot 5}$

Step 1.4.1.5

Multiply $- 20$ by $- 3$ .

$y = \frac{1 \pm \sqrt{1 - 80 x + 60}}{2 \cdot 5}$

Step 1.4.1.6

Add $1$ and $60$ .

$y = \frac{1 \pm \sqrt{- 80 x + 61}}{2 \cdot 5}$

Step 1.4.2

Multiply $2$ by $5$ .

$y = \frac{1 \pm \sqrt{- 80 x + 61}}{10}$

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

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

$y = \frac{1 \pm \sqrt{1 - 4 \cdot 5 \cdot (4 x - 3)}}{2 \cdot 5}$

Step 1.5.1.2

Multiply $- 4$ by $5$ .

$y = \frac{1 \pm \sqrt{1 - 20 \cdot (4 x - 3)}}{2 \cdot 5}$

Step 1.5.1.3

Apply the distributive property.

$y = \frac{1 \pm \sqrt{1 - 20 (4 x) - 20 \cdot - 3}}{2 \cdot 5}$

Step 1.5.1.4

Multiply $4$ by $- 20$ .

$y = \frac{1 \pm \sqrt{1 - 80 x - 20 \cdot - 3}}{2 \cdot 5}$

Step 1.5.1.5

Multiply $- 20$ by $- 3$ .

$y = \frac{1 \pm \sqrt{1 - 80 x + 60}}{2 \cdot 5}$

Step 1.5.1.6

Add $1$ and $60$ .

$y = \frac{1 \pm \sqrt{- 80 x + 61}}{2 \cdot 5}$

Step 1.5.2

Multiply $2$ by $5$ .

$y = \frac{1 \pm \sqrt{- 80 x + 61}}{10}$

Step 1.5.3

Change the $\pm$ to $+$ .

$y = \frac{1 + \sqrt{- 80 x + 61}}{10}$

Step 1.6

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

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Step 1.6.1

Simplify the numerator.

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Step 1.6.1.1

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

$y = \frac{1 \pm \sqrt{1 - 4 \cdot 5 \cdot (4 x - 3)}}{2 \cdot 5}$

Step 1.6.1.2

Multiply $- 4$ by $5$ .

$y = \frac{1 \pm \sqrt{1 - 20 \cdot (4 x - 3)}}{2 \cdot 5}$

Step 1.6.1.3

Apply the distributive property.

$y = \frac{1 \pm \sqrt{1 - 20 (4 x) - 20 \cdot - 3}}{2 \cdot 5}$

Step 1.6.1.4

Multiply $4$ by $- 20$ .

$y = \frac{1 \pm \sqrt{1 - 80 x - 20 \cdot - 3}}{2 \cdot 5}$

Step 1.6.1.5

Multiply $- 20$ by $- 3$ .

$y = \frac{1 \pm \sqrt{1 - 80 x + 60}}{2 \cdot 5}$

Step 1.6.1.6

Add $1$ and $60$ .

$y = \frac{1 \pm \sqrt{- 80 x + 61}}{2 \cdot 5}$

Step 1.6.2

Multiply $2$ by $5$ .

$y = \frac{1 \pm \sqrt{- 80 x + 61}}{10}$

Step 1.6.3

Change the $\pm$ to $-$ .

$y = \frac{1 - \sqrt{- 80 x + 61}}{10}$

Step 1.7

The final answer is the combination of both solutions.

$y = \frac{1 + \sqrt{- 80 x + 61}}{10}$

$y = \frac{1 - \sqrt{- 80 x + 61}}{10}$

$y = \frac{1 + \sqrt{- 80 x + 61}}{10}$

$y = \frac{1 - \sqrt{- 80 x + 61}}{10}$

Step 2

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

$y = \frac{1 + \sqrt{- 80 x + 61}}{10} \to f (x) = \frac{1 + \sqrt{- 80 x + 61}}{10}$

$y = \frac{1 - \sqrt{- 80 x + 61}}{10} \to f (x) = \frac{1 - \sqrt{- 80 x + 61}}{10}$

Step 3

Because the $y$ variable in the equation $4 x + 5 y^{2} - y = 3$ 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.

$\frac{d}{d x} (4 x + 5 y^{2} - y) = \frac{d}{d x} (3)$

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 $4 x + 5 y^{2} - y$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [5 y^{2}] + \frac{d}{d x} [- y]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [5 y^{2}] + \frac{d}{d x} [- y]$

Step 3.2.2

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

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Step 3.2.2.1

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

$4 \frac{d}{d x} [x] + \frac{d}{d x} [5 y^{2}] + \frac{d}{d x} [- y]$

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 $n = 1$ .

$4 \cdot 1 + \frac{d}{d x} [5 y^{2}] + \frac{d}{d x} [- y]$

Step 3.2.2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [5 y^{2}] + \frac{d}{d x} [- y]$

Step 3.2.3

Evaluate $\frac{d}{d x} [5 y^{2}]$ .

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Step 3.2.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 y^{2}$ with respect to $x$ is $5 \frac{d}{d x} [y^{2}]$ .

$4 + 5 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- y]$

Step 3.2.3.2

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 3.2.3.2.1

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

$4 + 5 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [- y]$

Step 3.2.3.2.2

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

$4 + 5 (2 u \frac{d}{d x} [y]) + \frac{d}{d x} [- y]$

Step 3.2.3.2.3

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

$4 + 5 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [- y]$

Step 3.2.3.3

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

$4 + 5 (2 y y') + \frac{d}{d x} [- y]$

Step 3.2.3.4

Multiply $2$ by $5$ .

$4 + 10 y y' + \frac{d}{d x} [- y]$

Step 3.2.4

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

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Step 3.2.4.1

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

$4 + 10 y y' - \frac{d}{d x} [y]$

Step 3.2.4.2

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

$4 + 10 y y' - y'$

Step 3.2.5

Reorder terms.

$10 y y' + 4 - y'$

Step 3.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0$

Step 3.4

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

$10 y y' + 4 - y' = 0$

Step 3.5

Solve for $y'$ .

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Step 3.5.1

Subtract $4$ from both sides of the equation.

$10 y y' - y' = - 4$

Step 3.5.2

Factor $y'$ out of $10 y y' - y'$ .

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Step 3.5.2.1

Factor $y'$ out of $10 y y'$ .

$y' (10 y) - y' = - 4$

Step 3.5.2.2

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

$y' (10 y) + y' \cdot - 1 = - 4$

Step 3.5.2.3

Factor $y'$ out of $y' (10 y) + y' \cdot - 1$ .

$y' (10 y - 1) = - 4$

Step 3.5.3

Divide each term in $y' (10 y - 1) = - 4$ by $10 y - 1$ and simplify.

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Step 3.5.3.1

Divide each term in $y' (10 y - 1) = - 4$ by $10 y - 1$ .

$\frac{y' (10 y - 1)}{10 y - 1} = \frac{- 4}{10 y - 1}$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $10 y - 1$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{y' (10 y - 1)}{10 y - 1} = \frac{- 4}{10 y - 1}$

Step 3.5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 4}{10 y - 1}$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Move the negative in front of the fraction.

$y' = - \frac{4}{10 y - 1}$

Step 3.6

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

$\frac{d y}{d x} = - \frac{4}{10 y - 1}$

Step 4

Set the derivative equal to $0$ then solve the equation $- \frac{4}{10 y - 1} = 0$ .

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Step 4.1

Set the numerator equal to zero.

$4 = 0$

Step 4.2

Since $4 \neq 0$ , there are no solutions.

No solution

Step 5

There are no solution found by setting the derivative equal to $0$ , $- \frac{4}{10 y - 1} = 0$ so there are no horizontal tangent lines.

No horizontal tangent lines found

Step 6