Calculus Examples

f (x) = 8 ln (x)

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

x = 1

$x = 1$

Step 1

Find the corresponding

y

$y$ -value to

x = 1

$x = 1$ .

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Step 1.1

Substitute

1

$1$ in for

x

$x$ .

y = 8 ln (1)

$y = 8 \ln (1)$

Step 1.2

Solve for

y

$y$ .

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Step 1.2.1

Remove parentheses.

y = 8 ln (1)

$y = 8 \ln (1)$

Step 1.2.2

Simplify

8 ln (1)

$8 \ln (1)$ .

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Step 1.2.2.1

The natural logarithm of

1

$1$ is

0

$0$ .

y = 8 \cdot 0

$y = 8 \cdot 0$

Step 1.2.2.2

Multiply

8

$8$ by

0

$0$ .

y = 0

$y = 0$

y = 0

$y = 0$

y = 0

$y = 0$

y = 0

$y = 0$

Step 2

Find the first derivative and evaluate at

x = 1

$x = 1$ and

y = 0

$y = 0$ to find the slope of the tangent line.

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Step 2.1

Since

8

$8$ is constant with respect to

x

$x$ , the derivative of

8 ln (x)

$8 \ln (x)$ with respect to

x

$x$ is

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

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

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

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

Step 2.2

The derivative of

ln (x)

$\ln (x)$ with respect to

x

$x$ is

\frac{1}{x}

$\frac{1}{x}$ .

8 \frac{1}{x}

$8 \frac{1}{x}$

Step 2.3

Combine

8

$8$ and

\frac{1}{x}

$\frac{1}{x}$ .

\frac{8}{x}

$\frac{8}{x}$

Step 2.4

Evaluate the derivative at

x = 1

$x = 1$ .

\frac{8}{1}

$\frac{8}{1}$

Step 2.5

Divide

8

$8$ by

1

$1$ .

8

$8$

8

$8$

Step 3

Plug the slope and point values into the point-slope formula and solve for

y

$y$ .

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Step 3.1

Use the slope

8

$8$ and a given point

(1, 0)

$(1, 0)$ to substitute for

x_{1}

$x_{1}$ and

y_{1}

$y_{1}$ in the point-slope form

y - y_{1} = m (x - x_{1})

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

y - (0) = 8 \cdot (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)

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

Step 3.3

Solve for

y

$y$ .

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Step 3.3.1

Add

y

$y$ and

0

$0$ .

y = 8 \cdot (x - 1)

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

Step 3.3.2

Simplify

8 \cdot (x - 1)

$8 \cdot (x - 1)$ .

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Step 3.3.2.1

Apply the distributive property.

y = 8 x + 8 \cdot - 1

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

Step 3.3.2.2

Multiply

8

$8$ by

- 1

$- 1$ .

y = 8 x - 8

$y = 8 x - 8$

y = 8 x - 8

$y = 8 x - 8$

y = 8 x - 8

$y = 8 x - 8$

y = 8 x - 8

$y = 8 x - 8$

Step 4