# What is the equation of the line that is normal to f(x)= lnx^2-x at  x= 1 ?

Mar 20, 2016

$y = - x$

#### Explanation:

A normal line is simply a line perpendicular to a tangent line. We are being asked to find the normal line at $x = 1$. In order to do that, we take the derivative, evaluate it at at $x = 1$ (which gives us the slope at $x = 1$), then use that information and a point on the line to find the normal line.

Step 1: Find the Derivative
Note first that $\ln {x}^{2}$ can be rewritten using the properties of logs to $2 \ln x$. Taking the derivative now is extremely easy: the derivative of $\ln x$ is $\frac{1}{x}$, which means the derivative of $2 \ln x = \frac{2}{x}$. As for $- x$, well, the derivative of that is just $- 1$. Applying this to the problem:
$f ' \left(x\right) = \frac{2}{x} - 1$
And that's all for this step.

Step 2: Evaluate
Here, we evaluate $f ' \left(1\right)$ to find the slope at $x = 1$:
$f ' \left(x\right) = \frac{2}{x} - 1$
$f ' \left(1\right) = \frac{2}{1} - 1 = 2 - 1 = 1$
But we don't want the slope of the tangent line, we want the slope of the normal line. Luckily, there is a simple relationship between tangent line and normal line slopes: they are opposite reciprocals. That is to say:
Normal line slope=-1/tangent line slope

In our case, that means the normal line slope is $- \frac{1}{1} = - 1$.

Step 3: Normal Line Equation
Normal lines, like tangent lines, are of the form $y = m x + b$, where $x$ and $y$ are points on the line, $m$ is the slope, and $b$ is the $y$-intercept. We have the slope ($- 1$), and we can easily get two points on the line. Using $x = 1$, we have:
$f \left(1\right) = 2 \ln \left(1\right) - \left(1\right) = 2 \left(0\right) - 1 = - 1$

Now we can solve for $b$:
$y = m x + b$
$- 1 = \left(1\right) \left(- 1\right) + b$
$- 1 = b - 1$
$b = 0$

The equation of the normal line is therefore $y = - x$.