Question

How do you find the equation of the tangent line to the graph of `y=ln(x)` at x=1?

Answer 1

Slope of tangent is `y=x-1`

Explanation:

At `x=1`, `y=ln1=0` i.e. curve passes through `(1,0)` at `x=1`

As `y=lnx`, `(dy)/(dx)=1/x` which gives slope of curve i.e. tangent at a point.

and hence at `x=1`, slope of tangent is `1/1=1`

Hence equation of line passing through `(1,0)` and slope `1` is

`(y-0)=1(x-1)` or

`y=x-1`