So we're trying to construct a line,
a tangent line to be more precise.
y = mx + b
The derivative evaluated at 2 gives us the SLOPE of that line.
f'(2) = m
Apply your quotient rule to get the derivative of f,
[tex]\large\rm \dfrac{(6x)'(2+x^2)-(6x)(2+x^2)'}{(2+x^2)^2}[/tex]
[tex]\large\rm \dfrac{(6)(2+x^2)-(6x)(2x)}{(2+x^2)^2}[/tex]
Don't bother simplifying.
Plug your location in at this point, it will be easier to simplify once everything is numbers:
[tex]\large\rm \dfrac{(6)(2+2^2)-(6\cdot2)(2\cdot2)}{(2+2^2)^2}[/tex]
which simplifies to something like 1/3.
This is the slope of our tangent line, m = 1/3.
y = (1/3)x + b
Since they gave you a point, and we just found our slope,
it might make more sense to put your line in Point-Slope form.
y - y0 = m(x - x0)
y - 2 = (1/3)(x - 2)
If you need your final answer in Slope-Intercept form,
then simply solve for y from this point.
