Calculus Examples

$4 x y^{3} + 5 x y = 36$ , $(4, 1)$

Step 1

Find the first derivative and evaluate at $x = 4$ and $y = 1$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (4 x y^{3} + 5 x y) = \frac{d}{d x} (36)$

Step 1.2

Differentiate the left side of the equation.

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Step 1.2.1

By the Sum Rule, the derivative of $4 x y^{3} + 5 x y$ with respect to $x$ is $\frac{d}{d x} [4 x y^{3}] + \frac{d}{d x} [5 x y]$ .

$\frac{d}{d x} [4 x y^{3}] + \frac{d}{d x} [5 x y]$

Step 1.2.2

Evaluate $\frac{d}{d x} [4 x y^{3}]$ .

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Step 1.2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x y^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x y^{3}]$ .

$4 \frac{d}{d x} [x y^{3}] + \frac{d}{d x} [5 x y]$

Step 1.2.2.2

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

$4 (x \frac{d}{d x} [y^{3}] + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 x y]$

Step 1.2.2.3

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 1.2.2.3.1

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

$4 (x (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 x y]$

Step 1.2.2.3.2

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

$4 (x (3 u^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 x y]$

Step 1.2.2.3.3

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

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

Step 1.2.2.4

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

$4 (x (3 y^{2} y') + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 x y]$

Step 1.2.2.5

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

$4 (x (3 y^{2} y') + y^{3} \cdot 1) + \frac{d}{d x} [5 x y]$

Step 1.2.2.6

Move $3$ to the left of $x$ .

$4 (3 \cdot x y^{2} y' + y^{3} \cdot 1) + \frac{d}{d x} [5 x y]$

Step 1.2.2.7

Multiply $y^{3}$ by $1$ .

$4 (3 x y^{2} y' + y^{3}) + \frac{d}{d x} [5 x y]$

Step 1.2.3

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

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Step 1.2.3.1

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

$4 (3 x y^{2} y' + y^{3}) + 5 \frac{d}{d x} [x y]$

Step 1.2.3.2

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 $g (x) = y$ .

$4 (3 x y^{2} y' + y^{3}) + 5 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 1.2.3.3

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

$4 (3 x y^{2} y' + y^{3}) + 5 (x y' + y \frac{d}{d x} [x])$

Step 1.2.3.4

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

$4 (3 x y^{2} y' + y^{3}) + 5 (x y' + y \cdot 1)$

Step 1.2.3.5

Multiply $y$ by $1$ .

$4 (3 x y^{2} y' + y^{3}) + 5 (x y' + y)$

Step 1.2.4

Simplify.

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Step 1.2.4.1

Apply the distributive property.

$4 (3 x y^{2} y') + 4 y^{3} + 5 (x y' + y)$

Step 1.2.4.2

Apply the distributive property.

$4 (3 x y^{2} y') + 4 y^{3} + 5 (x y') + 5 y$

Step 1.2.4.3

Multiply $3$ by $4$ .

$12 x y^{2} y' + 4 y^{3} + 5 x y' + 5 y$

Step 1.2.4.4

Reorder terms.

$4 y^{3} + 12 y^{2} x y' + 5 x y' + 5 y$

Step 1.3

Since $36$ is constant with respect to $x$ , the derivative of $36$ with respect to $x$ is $0$ .

$0$

Step 1.4

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

$4 y^{3} + 12 y^{2} x y' + 5 x y' + 5 y = 0$

Step 1.5

Solve for $y'$ .

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Step 1.5.1

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

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Step 1.5.1.1

Subtract $4 y^{3}$ from both sides of the equation.

$12 y^{2} x y' + 5 x y' + 5 y = - 4 y^{3}$

Step 1.5.1.2

Subtract $5 y$ from both sides of the equation.

$12 y^{2} x y' + 5 x y' = - 4 y^{3} - 5 y$

Step 1.5.2

Factor $x y'$ out of $12 y^{2} x y' + 5 x y'$ .

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Step 1.5.2.1

Factor $x y'$ out of $12 y^{2} x y'$ .

$x y' (12 y^{2}) + 5 x y' = - 4 y^{3} - 5 y$

Step 1.5.2.2

Factor $x y'$ out of $5 x y'$ .

$x y' (12 y^{2}) + x y' \cdot 5 = - 4 y^{3} - 5 y$

Step 1.5.2.3

Factor $x y'$ out of $x y' (12 y^{2}) + x y' \cdot 5$ .

$x y' (12 y^{2} + 5) = - 4 y^{3} - 5 y$

Step 1.5.3

Divide each term in $x y' (12 y^{2} + 5) = - 4 y^{3} - 5 y$ by $x (12 y^{2} + 5)$ and simplify.

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Step 1.5.3.1

Divide each term in $x y' (12 y^{2} + 5) = - 4 y^{3} - 5 y$ by $x (12 y^{2} + 5)$ .

$\frac{x y' (12 y^{2} + 5)}{x (12 y^{2} + 5)} = \frac{- 4 y^{3}}{x (12 y^{2} + 5)} + \frac{- 5 y}{x (12 y^{2} + 5)}$

Step 1.5.3.2

Simplify the left side.

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Step 1.5.3.2.1

Cancel the common factor of $x$ .

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Step 1.5.3.2.1.1

Cancel the common factor.

$\frac{x y' (12 y^{2} + 5)}{x (12 y^{2} + 5)} = \frac{- 4 y^{3}}{x (12 y^{2} + 5)} + \frac{- 5 y}{x (12 y^{2} + 5)}$

Step 1.5.3.2.1.2

Rewrite the expression.

$\frac{y' (12 y^{2} + 5)}{12 y^{2} + 5} = \frac{- 4 y^{3}}{x (12 y^{2} + 5)} + \frac{- 5 y}{x (12 y^{2} + 5)}$

Step 1.5.3.2.2

Cancel the common factor of $12 y^{2} + 5$ .

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Step 1.5.3.2.2.1

Cancel the common factor.

$\frac{y' (12 y^{2} + 5)}{12 y^{2} + 5} = \frac{- 4 y^{3}}{x (12 y^{2} + 5)} + \frac{- 5 y}{x (12 y^{2} + 5)}$

Step 1.5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 4 y^{3}}{x (12 y^{2} + 5)} + \frac{- 5 y}{x (12 y^{2} + 5)}$

Step 1.5.3.3

Simplify the right side.

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Step 1.5.3.3.1

Simplify each term.

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Step 1.5.3.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{4 y^{3}}{x (12 y^{2} + 5)} + \frac{- 5 y}{x (12 y^{2} + 5)}$

Step 1.5.3.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{4 y^{3}}{x (12 y^{2} + 5)} - \frac{5 y}{x (12 y^{2} + 5)}$

Step 1.6

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

$\frac{d y}{d x} = - \frac{4 y^{3}}{x (12 y^{2} + 5)} - \frac{5 y}{x (12 y^{2} + 5)}$

Step 1.7

Evaluate at $x = 4$ and $y = 1$ .

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Step 1.7.1

Replace the variable $x$ with $4$ in the expression.

$- \frac{4 y^{3}}{(4) (12 y^{2} + 5)} - \frac{5 y}{(4) (12 y^{2} + 5)}$

Step 1.7.2

Replace the variable $y$ with $1$ in the expression.

$- \frac{4 {(1)}^{3}}{(4) (12 {(1)}^{2} + 5)} - \frac{5 (1)}{(4) (12 {(1)}^{2} + 5)}$

Step 1.7.3

Simplify terms.

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Step 1.7.3.1

Combine the numerators over the common denominator.

$\frac{- 4 {(1)}^{3} - 5 \cdot 1}{4 (12 {(1)}^{2} + 5)}$

Step 1.7.3.2

Simplify each term.

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Step 1.7.3.2.1

One to any power is one.

$\frac{- 4 \cdot 1 - 5 \cdot 1}{4 (12 {(1)}^{2} + 5)}$

Step 1.7.3.2.2

Multiply $- 4$ by $1$ .

$\frac{- 4 - 5 \cdot 1}{4 (12 {(1)}^{2} + 5)}$

Step 1.7.3.2.3

Multiply $- 5$ by $1$ .

$\frac{- 4 - 5}{4 (12 {(1)}^{2} + 5)}$

Step 1.7.3.3

Subtract $5$ from $- 4$ .

$\frac{- 9}{4 (12 {(1)}^{2} + 5)}$

Step 1.7.4

Simplify the denominator.

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Step 1.7.4.1

One to any power is one.

$\frac{- 9}{4 (12 \cdot 1 + 5)}$

Step 1.7.4.2

Multiply $12$ by $1$ .

$\frac{- 9}{4 (12 + 5)}$

Step 1.7.4.3

Add $12$ and $5$ .

$\frac{- 9}{4 \cdot 17}$

Step 1.7.5

Simplify the expression.

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Step 1.7.5.1

Multiply $4$ by $17$ .

$\frac{- 9}{68}$

Step 1.7.5.2

Move the negative in front of the fraction.

$- \frac{9}{68}$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 2.1

Use the slope $- \frac{9}{68}$ and a given point $(4, 1)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = - \frac{9}{68} \cdot (x - (4))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y - 1 = - \frac{9}{68} \cdot (x - 4)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $- \frac{9}{68} \cdot (x - 4)$ .

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Step 2.3.1.1

Rewrite.

$y - 1 = 0 + 0 - \frac{9}{68} \cdot (x - 4)$

Step 2.3.1.2

Simplify terms.

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Step 2.3.1.2.1

Apply the distributive property.

$y - 1 = - \frac{9}{68} x - \frac{9}{68} \cdot - 4$

Step 2.3.1.2.2

Combine $x$ and $\frac{9}{68}$ .

$y - 1 = - \frac{x \cdot 9}{68} - \frac{9}{68} \cdot - 4$

Step 2.3.1.2.3

Cancel the common factor of $4$ .

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Step 2.3.1.2.3.1

Move the leading negative in $- \frac{9}{68}$ into the numerator.

$y - 1 = - \frac{x \cdot 9}{68} + \frac{- 9}{68} \cdot - 4$

Step 2.3.1.2.3.2

Factor $4$ out of $68$ .

$y - 1 = - \frac{x \cdot 9}{68} + \frac{- 9}{4 (17)} \cdot - 4$

Step 2.3.1.2.3.3

Factor $4$ out of $- 4$ .

$y - 1 = - \frac{x \cdot 9}{68} + \frac{- 9}{4 \cdot 17} \cdot (4 \cdot - 1)$

Step 2.3.1.2.3.4

Cancel the common factor.

$y - 1 = - \frac{x \cdot 9}{68} + \frac{- 9}{4 \cdot 17} \cdot (4 \cdot - 1)$

Step 2.3.1.2.3.5

Rewrite the expression.

$y - 1 = - \frac{x \cdot 9}{68} + \frac{- 9}{17} \cdot - 1$

Step 2.3.1.2.4

Combine $\frac{- 9}{17}$ and $- 1$ .

$y - 1 = - \frac{x \cdot 9}{68} + \frac{- 9 \cdot - 1}{17}$

Step 2.3.1.2.5

Multiply $- 9$ by $- 1$ .

$y - 1 = - \frac{x \cdot 9}{68} + \frac{9}{17}$

Step 2.3.1.3

Move $9$ to the left of $x$ .

$y - 1 = - \frac{9 x}{68} + \frac{9}{17}$

Step 2.3.2

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

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Step 2.3.2.1

Add $1$ to both sides of the equation.

$y = - \frac{9 x}{68} + \frac{9}{17} + 1$

Step 2.3.2.2

Write $1$ as a fraction with a common denominator.

$y = - \frac{9 x}{68} + \frac{9}{17} + \frac{17}{17}$

Step 2.3.2.3

Combine the numerators over the common denominator.

$y = - \frac{9 x}{68} + \frac{9 + 17}{17}$

Step 2.3.2.4

Add $9$ and $17$ .

$y = - \frac{9 x}{68} + \frac{26}{17}$

Step 2.3.3

Write in $y = m x + b$ form.

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Step 2.3.3.1

Reorder terms.

$y = - (\frac{9}{68} x) + \frac{26}{17}$

Step 2.3.3.2

Remove parentheses.

$y = - \frac{9}{68} x + \frac{26}{17}$

Step 3