Calculus Examples

$f (x) = 8 \ln (x)$ at $x = 1$

Step 1

Find the corresponding $y$ -value to $x = 1$ .

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Step 1.1

Substitute $1$ in for $x$ .

$y = 8 \ln (1)$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = 8 \ln (1)$

Step 1.2.2

Simplify $8 \ln (1)$ .

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Step 1.2.2.1

The natural logarithm of $1$ is $0$ .

$y = 8 \cdot 0$

Step 1.2.2.2

Multiply $8$ by $0$ .

$y = 0$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = 0$ to find the slope of the tangent line.

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Step 2.1

Since $8$ is constant with respect to $x$ , the derivative of $8 \ln (x)$ with respect to $x$ is $8 \frac{d}{d x} [\ln (x)]$ .

$8 \frac{d}{d x} [\ln (x)]$

Step 2.2

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

$8 \frac{1}{x}$

Step 2.3

Combine $8$ and $\frac{1}{x}$ .

$\frac{8}{x}$

Step 2.4

Evaluate the derivative at $x = 1$ .

$\frac{8}{1}$

Step 2.5

Divide $8$ by $1$ .

$8$

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 $8$ and a given point $(1, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = 8 \cdot (x - (1))$

Step 3.2

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

$y + 0 = 8 \cdot (x - 1)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Add $y$ and $0$ .

$y = 8 \cdot (x - 1)$

Step 3.3.2

Simplify $8 \cdot (x - 1)$ .

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Step 3.3.2.1

Apply the distributive property.

$y = 8 x + 8 \cdot - 1$

Step 3.3.2.2

Multiply $8$ by $- 1$ .

$y = 8 x - 8$

Step 4