Question

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

Answer 1

`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 `lnx^2` can be rewritten using the properties of logs to `2lnx`. Taking the derivative now is extremely easy: the derivative of `lnx` is `1/x`, which means the derivative of `2lnx=2/x`. As for `-x`, well, the derivative of that is just `-1`. Applying this to the problem:
`f'(x)=2/x-1`
And that's all for this step.

Step 2: Evaluate
Here, we evaluate `f'(1)` to find the slope at `x=1`:
`f'(x)=2/x-1`
`f'(1)=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 `-1/1=-1`.

Step 3: Normal Line Equation
Normal lines, like tangent lines, are of the form `y=mx+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(1)=2ln(1)-(1)=2(0)-1=-1`

Now we can solve for `b`:
`y=mx+b`
`-1=(1)(-1)+b`
`-1=b-1`
`b=0`

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