Need an answer to this please (derivative)?

Question

given that f(x)=x/3x^2+3, determine the tangent line at the point (2, 2/15)

Note that the +3 ISN'T in the exponent, just the 2

According to the program I'm using that's not the right answer, can someone else help?

Answer 1

Use the quotient rule to find the deriv:

f'(x) = [(3x^2 + 3)(1) - x(6x)]\(3x^2 + 3)^2

Then plug in 2 for x:

slope = [15-24]/15^2 = -9/225 = -1/25

Using the point-slope formula:

eqn tangent is

y - (2/15) = (-1/25)(x-2) **********

If you have to simplify:

y = (-1/25)x + 2/25 + 2/15

y = -1/25)x + 16/75

Answer 2

Are you sure you input the function into the program correctly?

All the work in the above answer is correct for the function x/(3x^2+3) at the given point.