Question

What is the equation of the tangent line of `f(x) =x-(ln(2x))^2` at `x = 3/2`?

Answer 1

`3y=(3-4ln3)x+3ln3(2-ln3)`

Explanation:

In order to find the equation of the tangent, we first need to find its gradient. We can do this be differentiating `f(x)` and evaluating the derivative at the point where the tangent touches the curve.

`f(x)=x-ln^2(2x)`

`f'(x)=1-(2ln2x)/x`

`f'(3/2)=1-(2ln3)/(3/2)=1-(4ln3)/3`

`thereforem=1-(4ln3)/3`

`f(3/2)=3/2-ln^2(3)`

`y-(3/2-ln^2 (3))=(1-(4ln3)/3)(x-3/2)`

`y-(3/2-ln^2 (3))=(1-(4ln3)/3)x+3/2((4ln3)/3-1)`

`y=(1-(4ln3)/3)x+2ln3-3/2+3/2-ln^2(3)`

`y=(3-4ln3)/3x+ln3(2-ln3)`

`3y=(3-4ln3)x+3ln3(2-ln3)`