Calculus Examples

$f (x) = \ln^{5} (x)$ at $x = 9$

Step 1

Find the corresponding $y$ -value to $x = 9$ .

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Step 1.1

Substitute $9$ in for $x$ .

$y = \ln ({(9)}^{5})$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = \ln (9^{5})$

Step 1.2.2

Remove parentheses.

$y = \ln ({(9)}^{5})$

Step 1.2.3

Raise $9$ to the power of $5$ .

$y = \ln (59049)$

Step 2

Find the first derivative and evaluate at $x = 9$ and $y = \ln (59049)$ to find the slope of the tangent line.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = {(x)}^{5}$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as ${(x)}^{5}$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [{(x)}^{5}]$

Step 2.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [{(x)}^{5}]$

Step 2.1.3

Replace all occurrences of $u$ with ${(x)}^{5}$ .

$\frac{1}{x^{5}} \frac{d}{d x} [x^{5}]$

Step 2.2

Differentiate using the Power Rule.

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Step 2.2.1

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{1}{x^{5}} (5 x^{4})$

Step 2.2.2

Simplify terms.

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Step 2.2.2.1

Combine $5$ and $\frac{1}{x^{5}}$ .

$\frac{5}{x^{5}} x^{4}$

Step 2.2.2.2

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

$\frac{5 x^{4}}{x^{5}}$

Step 2.2.2.3

Cancel the common factor of $x^{4}$ and $x^{5}$ .

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Step 2.2.2.3.1

Factor $x^{4}$ out of $5 x^{4}$ .

$\frac{x^{4} \cdot 5}{x^{5}}$

Step 2.2.2.3.2

Cancel the common factors.

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Step 2.2.2.3.2.1

Factor $x^{4}$ out of $x^{5}$ .

$\frac{x^{4} \cdot 5}{x^{4} x}$

Step 2.2.2.3.2.2

Cancel the common factor.

$\frac{x^{4} \cdot 5}{x^{4} x}$

Step 2.2.2.3.2.3

Rewrite the expression.

$\frac{5}{x}$

Step 2.3

Evaluate the derivative at $x = 9$ .

$\frac{5}{9}$

Step 3

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

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Step 3.1

Use the slope $\frac{5}{9}$ and a given point $(9, \ln (59049))$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (\ln (59049)) = \frac{5}{9} \cdot (x - (9))$

Step 3.2

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

$y - \ln (59049) = \frac{5}{9} \cdot (x - 9)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $\frac{5}{9} \cdot (x - 9)$ .

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Step 3.3.1.1

Rewrite.

$y - \ln (59049) = 0 + 0 + \frac{5}{9} \cdot (x - 9)$

Step 3.3.1.2

Simplify by adding zeros.

$y - \ln (59049) = \frac{5}{9} \cdot (x - 9)$

Step 3.3.1.3

Apply the distributive property.

$y - \ln (59049) = \frac{5}{9} x + \frac{5}{9} \cdot - 9$

Step 3.3.1.4

Combine $\frac{5}{9}$ and $x$ .

$y - \ln (59049) = \frac{5 x}{9} + \frac{5}{9} \cdot - 9$

Step 3.3.1.5

Cancel the common factor of $9$ .

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Step 3.3.1.5.1

Factor $9$ out of $- 9$ .

$y - \ln (59049) = \frac{5 x}{9} + \frac{5}{9} \cdot (9 (- 1))$

Step 3.3.1.5.2

Cancel the common factor.

$y - \ln (59049) = \frac{5 x}{9} + \frac{5}{9} \cdot (9 \cdot - 1)$

Step 3.3.1.5.3

Rewrite the expression.

$y - \ln (59049) = \frac{5 x}{9} + 5 \cdot - 1$

Step 3.3.1.6

Multiply $5$ by $- 1$ .

$y - \ln (59049) = \frac{5 x}{9} - 5$

Step 3.3.2

Add $\ln (59049)$ to both sides of the equation.

$y = \frac{5 x}{9} - 5 + \ln (59049)$

Step 3.3.3

Reorder terms.

$y = \frac{5}{9} x - 5 + \ln (59049)$

Step 4