Calculus Examples

$x^{2} + y^{2} - 2 x + 4 y = 8$ , $(4, 0)$

Step 1

Find the first derivative and evaluate at $x = 4$ and $y = 0$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} + y^{2} - 2 x + 4 y) = \frac{d}{d x} (8)$

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

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

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 $n = 2$ .

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

Step 1.2.2

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

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Step 1.2.2.1

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 1.2.2.1.1

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

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

Step 1.2.2.1.2

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

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

Step 1.2.2.1.3

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

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

Step 1.2.2.2

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

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

Step 1.2.3.2

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

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

Step 1.2.3.3

Multiply $- 2$ by $1$ .

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

Step 1.2.4

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

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Step 1.2.4.1

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

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

Step 1.2.4.2

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

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

Step 1.2.5

Reorder terms.

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

Step 1.3

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$0$

Step 1.4

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

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

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.

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

Step 1.5.1.2

Add $2$ to both sides of the equation.

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

Step 1.5.2

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

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Step 1.5.2.1

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

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

Step 1.5.2.2

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

$2 y' y + 2 y' \cdot 2 = - 2 x + 2$

Step 1.5.2.3

Factor $2 y'$ out of $2 y' y + 2 y' \cdot 2$ .

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

Step 1.5.3

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

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Step 1.5.3.1

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

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

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.

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

Step 1.5.3.2.1.2

Rewrite the expression.

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

Step 1.5.3.2.2

Cancel the common factor of $y + 2$ .

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Step 1.5.3.2.2.1

Cancel the common factor.

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

Step 1.5.3.2.2.2

Divide $y'$ by $1$ .

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

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 $- 2 x$ .

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

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.

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

Step 1.5.3.3.1.1.2.2

Rewrite the expression.

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

Step 1.5.3.3.1.2

Move the negative in front of the fraction.

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

Step 1.5.3.3.1.3

Cancel the common factor of $2$ .

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Step 1.5.3.3.1.3.1

Cancel the common factor.

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

Step 1.5.3.3.1.3.2

Rewrite the expression.

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

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.

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

Step 1.5.3.3.2.2

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

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

Step 1.5.3.3.2.3

Rewrite $1$ as $- 1 (- 1)$ .

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

Step 1.5.3.3.2.4

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

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

Step 1.5.3.3.2.5

Simplify the expression.

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Step 1.5.3.3.2.5.1

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

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

Step 1.5.3.3.2.5.2

Move the negative in front of the fraction.

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

Step 1.6

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

$\frac{d y}{d x} = - \frac{x - 1}{y + 2}$

Step 1.7

Evaluate at $x = 4$ and $y = 0$ .

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Step 1.7.1

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

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

Step 1.7.2

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

$- \frac{(4) - 1}{(0) + 2}$

Step 1.7.3

Subtract $1$ from $4$ .

$- \frac{3}{0 + 2}$

Step 1.7.4

Add $0$ and $2$ .

$- \frac{3}{2}$

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 $- \frac{3}{2}$ and a given point $(4, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = - \frac{3}{2} \cdot (x - (4))$

Step 2.2

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

$y + 0 = - \frac{3}{2} \cdot (x - 4)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

$y = - \frac{3}{2} \cdot (x - 4)$

Step 2.3.2

Simplify $- \frac{3}{2} \cdot (x - 4)$ .

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Step 2.3.2.1

Simplify terms.

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Step 2.3.2.1.1

Apply the distributive property.

$y = - \frac{3}{2} x - \frac{3}{2} \cdot - 4$

Step 2.3.2.1.2

Combine $x$ and $\frac{3}{2}$ .

$y = - \frac{x \cdot 3}{2} - \frac{3}{2} \cdot - 4$

Step 2.3.2.1.3

Cancel the common factor of $2$ .

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Step 2.3.2.1.3.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$y = - \frac{x \cdot 3}{2} + \frac{- 3}{2} \cdot - 4$

Step 2.3.2.1.3.2

Factor $2$ out of $- 4$ .

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

Step 2.3.2.1.3.3

Cancel the common factor.

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

Step 2.3.2.1.3.4

Rewrite the expression.

$y = - \frac{x \cdot 3}{2} - 3 \cdot - 2$

Step 2.3.2.1.4

Multiply $- 3$ by $- 2$ .

$y = - \frac{x \cdot 3}{2} + 6$

Step 2.3.2.2

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

$y = - \frac{3 x}{2} + 6$

Step 2.3.3

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

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Step 2.3.3.1

Reorder terms.

$y = - (\frac{3}{2} x) + 6$

Step 2.3.3.2

Remove parentheses.

$y = - \frac{3}{2} x + 6$

Step 3