Calculus Examples

$f (x) = 6 + \ln (x)$ ; $x = 1$

Step 1

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

Tap for more steps...

Step 1.1

Substitute $1$ in for $x$ .

$y = 6 + \ln (1)$

Step 1.2

Solve for $y$ .

Tap for more steps...

Step 1.2.1

Remove parentheses.

$y = 6 + \ln (1)$

Step 1.2.2

Simplify $6 + \ln (1)$ .

Tap for more steps...

Step 1.2.2.1

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

$y = 6 + 0$

Step 1.2.2.2

Add $6$ and $0$ .

$y = 6$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = 6$ to find the slope of the tangent line.

Tap for more steps...

Step 2.1

Differentiate.

Tap for more steps...

Step 2.1.1

By the Sum Rule, the derivative of $6 + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [6] + \frac{d}{d x} [\ln (x)]$ .

$\frac{d}{d x} [6] + \frac{d}{d x} [\ln (x)]$

Step 2.1.2

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

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

Step 2.2

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

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

Step 2.3

Add $0$ and $\frac{1}{x}$ .

$\frac{1}{x}$

Step 2.4

Evaluate the derivative at $x = 1$ .

$\frac{1}{1}$

Step 2.5

Divide $1$ by $1$ .

$1$

Step 3

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

Tap for more steps...

Step 3.1

Use the slope $1$ and a given point $(1, 6)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (6) = 1 \cdot (x - (1))$

Step 3.2

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

$y - 6 = 1 \cdot (x - 1)$

Step 3.3

Solve for $y$ .

Tap for more steps...

Step 3.3.1

Multiply $x - 1$ by $1$ .

$y - 6 = x - 1$

Step 3.3.2

Move all terms not containing $y$ to the right side of the equation.

Tap for more steps...

Step 3.3.2.1

Add $6$ to both sides of the equation.

$y = x - 1 + 6$

Step 3.3.2.2

Add $- 1$ and $6$ .

$y = x + 5$

Step 4