Calculus Examples

$f (x) = \frac{2}{x}$ at $(5, \frac{2}{5})$

Step 1

Find the first derivative and evaluate at $x = 5$ and $y = \frac{2}{5}$ to find the slope of the tangent line.

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Step 1.1

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{x}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{x}]$ .

$2 \frac{d}{d x} [\frac{1}{x}]$

Step 1.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$2 \frac{d}{d x} [x^{- 1}]$

Step 1.3

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

$2 (- x^{- 2})$

Step 1.4

Multiply $- 1$ by $2$ .

$- 2 x^{- 2}$

Step 1.5

Simplify.

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Step 1.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 2 \frac{1}{x^{2}}$

Step 1.5.2

Combine terms.

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Step 1.5.2.1

Combine $- 2$ and $\frac{1}{x^{2}}$ .

$\frac{- 2}{x^{2}}$

Step 1.5.2.2

Move the negative in front of the fraction.

$- \frac{2}{x^{2}}$

Step 1.6

Evaluate the derivative at $x = 5$ .

$- \frac{2}{{(5)}^{2}}$

Step 1.7

Raise $5$ to the power of $2$ .

$- \frac{2}{25}$

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 $- \frac{2}{25}$ and a given point $(5, \frac{2}{5})$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (\frac{2}{5}) = - \frac{2}{25} \cdot (x - (5))$

Step 2.2

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

$y - \frac{2}{5} = - \frac{2}{25} \cdot (x - 5)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $- \frac{2}{25} \cdot (x - 5)$ .

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Step 2.3.1.1

Rewrite.

$y - \frac{2}{5} = 0 + 0 - \frac{2}{25} \cdot (x - 5)$

Step 2.3.1.2

Simplify terms.

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Step 2.3.1.2.1

Apply the distributive property.

$y - \frac{2}{5} = - \frac{2}{25} x - \frac{2}{25} \cdot - 5$

Step 2.3.1.2.2

Combine $x$ and $\frac{2}{25}$ .

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} - \frac{2}{25} \cdot - 5$

Step 2.3.1.2.3

Cancel the common factor of $5$ .

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Step 2.3.1.2.3.1

Move the leading negative in $- \frac{2}{25}$ into the numerator.

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{- 2}{25} \cdot - 5$

Step 2.3.1.2.3.2

Factor $5$ out of $25$ .

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{- 2}{5 (5)} \cdot - 5$

Step 2.3.1.2.3.3

Factor $5$ out of $- 5$ .

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{- 2}{5 \cdot 5} \cdot (5 \cdot - 1)$

Step 2.3.1.2.3.4

Cancel the common factor.

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{- 2}{5 \cdot 5} \cdot (5 \cdot - 1)$

Step 2.3.1.2.3.5

Rewrite the expression.

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{- 2}{5} \cdot - 1$

Step 2.3.1.2.4

Combine $\frac{- 2}{5}$ and $- 1$ .

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{- 2 \cdot - 1}{5}$

Step 2.3.1.2.5

Multiply $- 2$ by $- 1$ .

$y - \frac{2}{5} = - \frac{x \cdot 2}{25} + \frac{2}{5}$

Step 2.3.1.3

Move $2$ to the left of $x$ .

$y - \frac{2}{5} = - \frac{2 x}{25} + \frac{2}{5}$

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 $\frac{2}{5}$ to both sides of the equation.

$y = - \frac{2 x}{25} + \frac{2}{5} + \frac{2}{5}$

Step 2.3.2.2

Combine the numerators over the common denominator.

$y = \frac{- 2 x}{25} + \frac{2 + 2}{5}$

Step 2.3.2.3

Add $2$ and $2$ .

$y = \frac{- 2 x}{25} + \frac{4}{5}$

Step 2.3.2.4

Move the negative in front of the fraction.

$y = - \frac{2 x}{25} + \frac{4}{5}$

Step 2.3.3

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

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Step 2.3.3.1

Reorder terms.

$y = - (\frac{2}{25} x) + \frac{4}{5}$

Step 2.3.3.2

Remove parentheses.

$y = - \frac{2}{25} x + \frac{4}{5}$

Step 3