Question

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

Answer 1

`y = (2/3ln9-6)x +ln^2 9 - 2ln9+9`

Explanation:

We have `f(x) = ln^2(3x)-x^2`

The gradient of the tangent at any particular point is given by the derivative at that point.

Differentiating wrt `x` (using the chain rule);

`\ \ \ \ \ f'(x) = 2ln(3x)*1/(3x)*3-2x`
`:. f'(x) = (2ln(3x))/x - 2x`

We need to find `f(x)` and `f'(x)` when `x=3`

`x=3 => f(3) \ \ = ln^2 9-9`
`x=3 => f'(3) = 2/3ln9-6`

So the tangent we seek passes through the point `(3, ln^2 9-9)` and has slope `m=2/3ln9-6`, so using `y-y_1=m(x-x_1)` the required equation is;

`y - (ln^2 9-9) = (2/3ln9-6)(x - 3)`
`:. y - (ln^2 9-9) = (2/3ln9-6)(x - 3)`
`:. y - ln^2 9 + 9 = (2/3ln9-6)x - 3(2/3ln9-6)`
`:. y - ln^2 9 + 9 = (2/3ln9-6)x - 2ln9+18`
`:. y = (2/3ln9-6)x +ln^2 9 - 2ln9+9`

We can confirm this solution is correct by looking at the graph: