Question

What is the equation of the normal line of `f(x)=x/lnx` at `x=4`?

Answer 1

`y = (ln(4)-1)/(ln(4)^2)(x-4)+4/(ln4)`

Explanation:

First of all, we need a point of intersection. We know the line will intersect `f(x)` at `(4,f(4))` so we will use to obtain the point.

`f(4) = 4/ln(4)`

So we know at least `(4, 4/ln4)` is a point on the line.

Next we need the gradient of the line and this is obtained by taking the derivative of `f(x)`. Using the quotient rule to differentiate we get:

`f'(x) = (ln(x) - 1)/ln(x)^2`

Now find `f'(4)=(ln(4)-1)/ln(4)^2`

Thus we have the gradient of the line.

Now substitute into the formula:

`y-b=m(x-a)` where `m` is the gradient and `(a,b)` is the known point of intersection:

`y-4/(ln4)=(ln(4)-1)/(ln(4)^2)(x-4)`

`y = (ln(4)-1)/(ln(4)^2)(x-4)+4/(ln4)`

We can better visualise this from the graphs of the function: