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

Nov 6, 2016

The normal line has equation $y = \frac{1}{2} x + \frac{1}{2}$.

#### Explanation:

Let's start by finding the corresponding $y$-coordinate that the function $f \left(x\right)$ passes through.

$f \left(1\right) = \frac{1}{1} - \ln \frac{1}{1}$

$f \left(1\right) = 1 - \frac{0}{1}$

$f \left(1\right) = 1$

Now, let's differentiate using the power rule, the quotient rule and the identity $\left(\ln x\right) ' = \frac{1}{x}$.

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

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

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

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

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

We now determine the slope of the tangent by inserting our point into the derivative (since this represents the instantaneous rate of change of the function).

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

$f ' \left(1\right) = \frac{- 2 + 0}{1}$

$f ' \left(1\right) = - 2$

Hence, the slope of the tangent is $- 2$. The normal line is always perpendicular to the tangent, so the slope of the normal line is the negative reciprocal of the tangent.

$\therefore$ The slope of the normal line is $\frac{1}{2}$.

We can now determine the equation of the normal line, because we know the slope $\left(1\right)$, and the point of contact $\left(1 , 1\right)$.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = \frac{1}{2} \left(x - 1\right)$

$y - 1 = \frac{1}{2} x - \frac{1}{2}$

$y = \frac{1}{2} x + \frac{1}{2}$

Hopefully this helps!