Calculus Examples

$2 x^{2} + y^{2} = 17$ , $(2, 3)$

Step 1

Find the first derivative and evaluate at $x = 2$ and $y = 3$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (2 x^{2} + y^{2}) = \frac{d}{d x} (17)$

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

Step 1.2.2.2

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

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

Step 1.2.2.3

Multiply $2$ by $2$ .

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

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 $y$ .

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

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 $n = 2$ .

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

Step 1.2.3.1.3

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

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

Step 1.2.3.2

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

$4 x + 2 y y'$

Step 1.2.4

Reorder terms.

$2 y y' + 4 x$

Step 1.3

Since $17$ is constant with respect to $x$ , the derivative of $17$ with respect to $x$ is $0$ .

$0$

Step 1.4

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

$2 y y' + 4 x = 0$

Step 1.5

Solve for $y'$ .

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Step 1.5.1

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

$2 y y' = - 4 x$

Step 1.5.2

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

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Step 1.5.2.1

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

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

Step 1.5.2.2

Simplify the left side.

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Step 1.5.2.2.1

Cancel the common factor of $2$ .

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Step 1.5.2.2.1.1

Cancel the common factor.

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

Step 1.5.2.2.1.2

Rewrite the expression.

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

Step 1.5.2.2.2

Cancel the common factor of $y$ .

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Step 1.5.2.2.2.1

Cancel the common factor.

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

Step 1.5.2.2.2.2

Divide $y'$ by $1$ .

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

Step 1.5.2.3

Simplify the right side.

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Step 1.5.2.3.1

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

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Step 1.5.2.3.1.1

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

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

Step 1.5.2.3.1.2

Cancel the common factors.

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Step 1.5.2.3.1.2.1

Factor $2$ out of $2 y$ .

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

Step 1.5.2.3.1.2.2

Cancel the common factor.

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

Step 1.5.2.3.1.2.3

Rewrite the expression.

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

Step 1.5.2.3.2

Move the negative in front of the fraction.

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

Step 1.6

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

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

Step 1.7

Evaluate at $x = 2$ and $y = 3$ .

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Step 1.7.1

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

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

Step 1.7.2

Replace the variable $y$ with $3$ in the expression.

$- \frac{2 (2)}{3}$

Step 1.7.3

Multiply $2$ by $2$ .

$- \frac{4}{3}$

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

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

Step 2.2

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

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

Step 2.3

Solve for $y$ .

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Step 2.3.1

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

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Step 2.3.1.1

Rewrite.

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

Step 2.3.1.2

Simplify by adding zeros.

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

Step 2.3.1.3

Apply the distributive property.

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

Step 2.3.1.4

Combine $x$ and $\frac{4}{3}$ .

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

Step 2.3.1.5

Multiply $- \frac{4}{3} \cdot - 2$ .

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Step 2.3.1.5.1

Multiply $- 2$ by $- 1$ .

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

Step 2.3.1.5.2

Combine $2$ and $\frac{4}{3}$ .

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

Step 2.3.1.5.3

Multiply $2$ by $4$ .

$y - 3 = - \frac{x \cdot 4}{3} + \frac{8}{3}$

Step 2.3.1.6

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

$y - 3 = - \frac{4 x}{3} + \frac{8}{3}$

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 $3$ to both sides of the equation.

$y = - \frac{4 x}{3} + \frac{8}{3} + 3$

Step 2.3.2.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = - \frac{4 x}{3} + \frac{8}{3} + 3 \cdot \frac{3}{3}$

Step 2.3.2.3

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

$y = - \frac{4 x}{3} + \frac{8}{3} + \frac{3 \cdot 3}{3}$

Step 2.3.2.4

Combine the numerators over the common denominator.

$y = - \frac{4 x}{3} + \frac{8 + 3 \cdot 3}{3}$

Step 2.3.2.5

Simplify the numerator.

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Step 2.3.2.5.1

Multiply $3$ by $3$ .

$y = - \frac{4 x}{3} + \frac{8 + 9}{3}$

Step 2.3.2.5.2

Add $8$ and $9$ .

$y = - \frac{4 x}{3} + \frac{17}{3}$

Step 2.3.3

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

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Step 2.3.3.1

Reorder terms.

$y = - (\frac{4}{3} x) + \frac{17}{3}$

Step 2.3.3.2

Remove parentheses.

$y = - \frac{4}{3} x + \frac{17}{3}$

Step 3