Question

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

Answer 1

`y=5x-25+ln(1/5)`

Explanation:

According to the Chain Rule,

`f'(x)=(d/dx[1/x])/(1/x)`

Let's find the derivative. (Remember that `1/x=x^-1`.)

`d/dx[x^-1]=-x^-2=-1/x^2`

Plug back in to find that:

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

We want the equation of the normal line at the point `(5,ln(1/5))`. 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'(5)=-1/5`

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

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

`y-ln(1/5)=5(x-5)`

Or, in slope-intercept form:

`y=5x-25+ln(1/5)`