# What is the equation of the normal line of f(x)=ln(1/x) at x=5?

Nov 20, 2015

$y = 5 x - 25 + \ln \left(\frac{1}{5}\right)$

#### Explanation:

According to the Chain Rule,

$f ' \left(x\right) = \frac{\frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]}{\frac{1}{x}}$

Let's find the derivative. (Remember that $\frac{1}{x} = {x}^{-} 1$.)

$\frac{d}{\mathrm{dx}} \left[{x}^{-} 1\right] = - {x}^{-} 2 = - \frac{1}{x} ^ 2$

Plug back in to find that:

$f ' \left(x\right) = \frac{- \frac{1}{x} ^ 2}{\frac{1}{x}} = - \frac{1}{x} ^ 2 \left(\frac{x}{1}\right) = - \frac{1}{x}$

We want the equation of the normal line at the point $\left(5 , \ln \left(\frac{1}{5}\right)\right)$. The normal line is perpendicular to the tangent line, so it has an opposite reciprocal slope.

We can find the slope of the tangent line at $x = 5$ by calculating $f ' \left(5\right) = - \frac{1}{5}$

Therefore, the slope of the normal line is $5$.

We can use point-slope form to write an equation:

$y - \ln \left(\frac{1}{5}\right) = 5 \left(x - 5\right)$

Or, in slope-intercept form:

$y = 5 x - 25 + \ln \left(\frac{1}{5}\right)$