Find the Tangent Line at x=1 f(x)=8+ natural log of x ; x=1

$f (x) = 8 + \ln (x)$ ; $x = 1$

Step 1

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

Substitute $1$ in for $x$ .

$y = 8 + \ln (1)$

Solve for $y$ .

Remove parentheses.

$y = 8 + \ln (1)$

Simplify $8 + \ln (1)$ .

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

$y = 8 + 0$

Add $8$ and $0$ .

$y = 8$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = 8$ to find the slope of the tangent line.

Differentiate.

By the Sum Rule, the derivative of $8 + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [\ln (x)]$ .

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

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

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

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

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

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

$\frac{1}{x}$

Evaluate the derivative at $x = 1$ .

$\frac{1}{1}$

Divide $1$ by $1$ .

$1$

Step 3

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

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

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

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

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

Solve for $y$ .

Multiply $x - 1$ by $1$ .

$y - 8 = x - 1$

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

Add $8$ to both sides of the equation.

$y = x - 1 + 8$

Add $- 1$ and $8$ .

$y = x + 7$

Step 4

Find the Tangent Line at x=1 f(x)=8+ natural log of x ; x=1

Related Questions

Support LCMGCF.com by disabling your adblocker.