Calculus Examples

$x^{2} + 4 x y + y^{2} = 13$ , $(2, 1)$

Step 1

Find the first derivative and evaluate at $x = 2$ and $y = 1$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} + 4 x y + y^{2}) = \frac{d}{d x} (13)$

Step 1.2

Differentiate the left side of the equation.

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Step 1.2.1

Differentiate.

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Step 1.2.1.1

By the Sum Rule, the derivative of $x^{2} + 4 x y + y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x y] + \frac{d}{d x} [y^{2}]$ .

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

Step 1.2.1.2

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

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 $g (x) = y$ .

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

Step 1.2.2.3

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

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

Step 1.2.2.4

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

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

Step 1.2.2.5

Multiply $y$ by $1$ .

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

Step 1.2.3

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

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Step 1.2.3.1

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 1.2.3.1.1

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

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

Step 1.2.3.1.2

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

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

Step 1.2.3.1.3

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

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

Step 1.2.3.2

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

$2 x + 4 (x y' + y) + 2 y y'$

Step 1.2.4

Simplify.

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Step 1.2.4.1

Apply the distributive property.

$2 x + 4 (x y') + 4 y + 2 y y'$

Step 1.2.4.2

Remove unnecessary parentheses.

$2 x + 4 x y' + 4 y + 2 y y'$

Step 1.2.4.3

Reorder terms.

$4 x y' + 2 y y' + 2 x + 4 y$

Step 1.3

Since $13$ is constant with respect to $x$ , the derivative of $13$ with respect to $x$ is $0$ .

$0$

Step 1.4

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

$4 x y' + 2 y y' + 2 x + 4 y = 0$

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 $2 x$ from both sides of the equation.

$4 x y' + 2 y y' + 4 y = - 2 x$

Step 1.5.1.2

Subtract $4 y$ from both sides of the equation.

$4 x y' + 2 y y' = - 2 x - 4 y$

Step 1.5.2

Factor $2 y'$ out of $4 x y' + 2 y y'$ .

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Step 1.5.2.1

Factor $2 y'$ out of $4 x y'$ .

$2 y' (2 x) + 2 y y' = - 2 x - 4 y$

Step 1.5.2.2

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

$2 y' (2 x) + 2 y' y = - 2 x - 4 y$

Step 1.5.2.3

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

$2 y' (2 x + y) = - 2 x - 4 y$

Step 1.5.3

Divide each term in $2 y' (2 x + y) = - 2 x - 4 y$ by $2 (2 x + y)$ and simplify.

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Step 1.5.3.1

Divide each term in $2 y' (2 x + y) = - 2 x - 4 y$ by $2 (2 x + y)$ .

$\frac{2 y' (2 x + y)}{2 (2 x + y)} = \frac{- 2 x}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.2

Simplify the left side.

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Step 1.5.3.2.1

Cancel the common factor of $2$ .

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Step 1.5.3.2.1.1

Cancel the common factor.

$\frac{2 y' (2 x + y)}{2 (2 x + y)} = \frac{- 2 x}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.2.1.2

Rewrite the expression.

$\frac{y' (2 x + y)}{2 x + y} = \frac{- 2 x}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.2.2

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

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Step 1.5.3.2.2.1

Cancel the common factor.

$\frac{y' (2 x + y)}{2 x + y} = \frac{- 2 x}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 2 x}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

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

Cancel the common factor of $- 2$ and $2$ .

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Step 1.5.3.3.1.1.1

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

$y' = \frac{2 (- x)}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.3.1.1.2

Cancel the common factors.

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Step 1.5.3.3.1.1.2.1

Cancel the common factor.

$y' = \frac{2 (- x)}{2 (2 x + y)} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.3.1.1.2.2

Rewrite the expression.

$y' = \frac{- x}{2 x + y} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{x}{2 x + y} + \frac{- 4 y}{2 (2 x + y)}$

Step 1.5.3.3.1.3

Cancel the common factor of $- 4$ and $2$ .

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Step 1.5.3.3.1.3.1

Factor $2$ out of $- 4 y$ .

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

Step 1.5.3.3.1.3.2

Cancel the common factors.

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Step 1.5.3.3.1.3.2.1

Cancel the common factor.

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

Step 1.5.3.3.1.3.2.2

Rewrite the expression.

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

Step 1.5.3.3.1.4

Move the negative in front of the fraction.

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

Step 1.5.3.3.2

Simplify terms.

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Step 1.5.3.3.2.1

Combine the numerators over the common denominator.

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

Step 1.5.3.3.2.2

Factor $- 1$ out of $- x$ .

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

Step 1.5.3.3.2.3

Factor $- 1$ out of $- 2 y$ .

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

Step 1.5.3.3.2.4

Factor $- 1$ out of $- (x) - (2 y)$ .

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

Step 1.5.3.3.2.5

Simplify the expression.

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Step 1.5.3.3.2.5.1

Rewrite $- (x + 2 y)$ as $- 1 (x + 2 y)$ .

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

Step 1.5.3.3.2.5.2

Move the negative in front of the fraction.

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

Step 1.6

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

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

Step 1.7

Evaluate at $x = 2$ and $y = 1$ .

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Step 1.7.1

Replace the variable $x$ with $2$ in the expression.

$- \frac{(2) + 2 y}{2 (2) + y}$

Step 1.7.2

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

$- \frac{(2) + 2 (1)}{2 (2) + 1}$

Step 1.7.3

Remove parentheses.

$- \frac{(2) + 2 (1)}{2 (2) + 1}$

Step 1.7.4

Simplify the numerator.

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Step 1.7.4.1

Multiply $2$ by $1$ .

$- \frac{2 + 2}{2 (2) + 1}$

Step 1.7.4.2

Add $2$ and $2$ .

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

Step 1.7.5

Simplify the denominator.

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Step 1.7.5.1

Multiply $2$ by $2$ .

$- \frac{4}{4 + 1}$

Step 1.7.5.2

Add $4$ and $1$ .

$- \frac{4}{5}$

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 $- \frac{4}{5}$ and a given point $(2, 1)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

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

Step 2.2

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

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

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $- \frac{4}{5} \cdot (x - 2)$ .

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Step 2.3.1.1

Rewrite.

$y - 1 = 0 + 0 - \frac{4}{5} \cdot (x - 2)$

Step 2.3.1.2

Simplify by adding zeros.

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

Step 2.3.1.3

Apply the distributive property.

$y - 1 = - \frac{4}{5} x - \frac{4}{5} \cdot - 2$

Step 2.3.1.4

Combine $x$ and $\frac{4}{5}$ .

$y - 1 = - \frac{x \cdot 4}{5} - \frac{4}{5} \cdot - 2$

Step 2.3.1.5

Multiply $- \frac{4}{5} \cdot - 2$ .

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Step 2.3.1.5.1

Multiply $- 2$ by $- 1$ .

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

Step 2.3.1.5.2

Combine $2$ and $\frac{4}{5}$ .

$y - 1 = - \frac{x \cdot 4}{5} + \frac{2 \cdot 4}{5}$

Step 2.3.1.5.3

Multiply $2$ by $4$ .

$y - 1 = - \frac{x \cdot 4}{5} + \frac{8}{5}$

Step 2.3.1.6

Move $4$ to the left of $x$ .

$y - 1 = - \frac{4 x}{5} + \frac{8}{5}$

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.

$y = - \frac{4 x}{5} + \frac{8}{5} + 1$

Step 2.3.2.2

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

$y = - \frac{4 x}{5} + \frac{8}{5} + \frac{5}{5}$

Step 2.3.2.3

Combine the numerators over the common denominator.

$y = - \frac{4 x}{5} + \frac{8 + 5}{5}$

Step 2.3.2.4

Add $8$ and $5$ .

$y = - \frac{4 x}{5} + \frac{13}{5}$

Step 2.3.3

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

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Step 2.3.3.1

Reorder terms.

$y = - (\frac{4}{5} x) + \frac{13}{5}$

Step 2.3.3.2

Remove parentheses.

$y = - \frac{4}{5} x + \frac{13}{5}$

Step 3