Calculus Examples

f (x) = 8 - ln (x)

$f (x) = 8 - \ln (x)$ ;

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

Simplify each term.

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Step 1.2.2.1.1

The natural logarithm of

1

$1$ is

0

$0$ .

y = 8 - 0

$y = 8 - 0$

Step 1.2.2.1.2

Multiply

- 1

$- 1$ by

0

$0$ .

y = 8 + 0

$y = 8 + 0$

y = 8 + 0

$y = 8 + 0$

Step 1.2.2.2

Add

8

$8$ and

0

$0$ .

y = 8

$y = 8$

y = 8

$y = 8$

y = 8

$y = 8$

y = 8

$y = 8$

Step 2

Find the first derivative and evaluate at

x = 1

$x = 1$ and

y = 8

$y = 8$ to find the slope of the tangent line.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of

8 - ln (x)

$8 - \ln (x)$ with respect to

x

$x$ is

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

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

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

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

Step 2.1.2

Since

8

$8$ is constant with respect to

x

$x$ , the derivative of

8

$8$ with respect to

x

$x$ is

0

$0$ .

0 + \frac{d}{d x} [- ln (x)]

$0 + \frac{d}{d x} [- \ln (x)]$

0 + \frac{d}{d x} [- ln (x)]

$0 + \frac{d}{d x} [- \ln (x)]$

Step 2.2

Evaluate

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

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

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Step 2.2.1

Since

- 1

$- 1$ is constant with respect to

x

$x$ , the derivative of

- ln (x)

$- \ln (x)$ with respect to

x

$x$ is

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

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

0 - \frac{d}{d x} [ln (x)]

$0 - \frac{d}{d x} [\ln (x)]$

Step 2.2.2

The derivative of

ln (x)

$\ln (x)$ with respect to

x

$x$ is

\frac{1}{x}

$\frac{1}{x}$ .

0 - \frac{1}{x}

$0 - \frac{1}{x}$

0 - \frac{1}{x}

$0 - \frac{1}{x}$

Step 2.3

Subtract

\frac{1}{x}

$\frac{1}{x}$ from

0

$0$ .

- \frac{1}{x}

$- \frac{1}{x}$

Step 2.4

Evaluate the derivative at

x = 1

$x = 1$ .

- \frac{1}{1}

$- \frac{1}{1}$

Step 2.5

Simplify.

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Step 2.5.1

Cancel the common factor of

1

$1$ .

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Step 2.5.1.1

Cancel the common factor.

$- \frac{1}{1}$

Step 2.5.1.2

Rewrite the expression.

$- 1 \cdot 1$

$- 1 \cdot 1$

Step 2.5.2

Multiply

$- 1$ by

$1$ .

$- 1$

$- 1$

$- 1$

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

$- 1$ and a given point

$(1, 8)$ 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

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

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

Step 3.2

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

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

Step 3.3

Solve for

$y$ .

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Step 3.3.1

Simplify

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

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Step 3.3.1.1

Rewrite.

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

Step 3.3.1.2

Simplify by adding zeros.

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

Step 3.3.1.3

Apply the distributive property.

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

Step 3.3.1.4

Simplify the expression.

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Step 3.3.1.4.1

Rewrite

$- 1 x$ as

$- x$ .

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

Step 3.3.1.4.2

Multiply

$- 1$ by

$- 1$ .

$y - 8 = - x + 1$

$y - 8 = - x + 1$

$y - 8 = - x + 1$

Step 3.3.2

Move all terms not containing

$y$ to the right side of the equation.

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Step 3.3.2.1

Add

$8$ to both sides of the equation.

$y = - x + 1 + 8$

Step 3.3.2.2

Add

$1$ and

$8$ .

$y = - x + 9$

$y = - x + 9$

$y = - x + 9$

$y = - x + 9$

Step 4

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$