Finding the derivative of the function and the slope of the tangent

[meeklobraca]

1. The problem statement, all variables and given/known data

Find the slope of the tangent to y=3x2 - 6x at x = 2 by first determining the derivative of the function from first principles


2. Relevant equations

f(x+h)-f(x) / h



3. The attempt at a solution

For the derivative I got -6x and the slope of the tangent is -12. Would this be correct?

Thanks!

[Dr. Lady]

You should check you're algebra in your equation... which should have a limit, of course.

[cosmic_tears]

no, your answer is not correct.
you obviously know how to find the derivative using the known formula (and not definition), and you can easily calculate y' to see that y'=6x-6.
So that should be the answer you get. Try again, it's an easy limit.

[meeklobraca]

Yup your right,

I got for the deriviative 6x - 6, with the slope being 6 at x = 2 ? Correct?

In my calculations I didnt get the -6 part of the deriviative cause i didnt account for the x with a zero exponent.

[Mark44]

As Dr. Lady points out, you should be doing this problem with a limit, and by your later question, I suspect that you are not doing it this way. That's what is meant by "first principles." If you're asked to find the derivative by first principles, and you don't use the definition of the limit, you are not likely to get full (or even partial) credit for your work.

[meeklobraca]

Okay I see your point. I used the definition of the limit in terms of finding the derivative. Which I used the lim = f(x+h) - f(x) / h formula.

And i fact using that formula I got -6x+6 for the derivative with the slope at x = 2 being -6. SO im a little confused at where I mixed the two up.

[Mark44]

For f'(x) you should not have gotten -6x + 6.

(f(x + h) - f(x))/h = [3(x + h)^2 -6(x + h) - (3x^2 - 6x)]/h
= [3x^2 + 6xh + 3h^2 -6x -6h -3x^2 + 6x]/h
[6xh + 3h^2 - 6h]/h

Now, factor h from each term in th numerator, and then take the limit as h approaches 0. That will give you f'(x).
After you have that, calculate f'(2).

[meeklobraca]

Yes thank you very much I see it now. My calculation error was in the -6 (x+h) area. Thanks!