Question

How do you find the equation of the tangent line to the graph of `f(x)= (ln x)^2` at x=6?

Answer 1

`y-(ln6)^2=ln6/3(x-6)`

Explanation:

At `x=6`, we have `f(x)=(ln6)^2`, hence we are seeking tangent at point `(6,(ln6)^2)`

Now slope of tangent is given by slope of curve at that point.

As `f'(x)=(2lnx)/x` and `f'(6)=ln6/3`

As tangent passes through `(6,(ln6)^2)` and has a slope of `ln6/3`, its equation is

`y-(ln6)^2=ln6/3(x-6)`

graph{(y-(ln6)^2-ln6/3(x-6))(y-(lnx)^2)=0 [-1.08, 18.92, -1.16, 8.84]}