## anonymous one year ago position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.

1. geerky42

$\large v(t) = \dfrac{\mathrm d }{\mathrm dt}~s(t)$Right?

2. geerky42

You just need to evaluate $$v(2)$$

3. anonymous

If I evaluate the formula at t=-2 I just get -14

4. anonymous

*2

5. geerky42

How did you get -14? Can you find derivative of -2 - 6t ?

6. anonymous

I don't think I know how to find derivatives properly.

7. anonymous

When it says derivative, I think derived from t equaling 2 and the formula -2 -6t. -2-6*(2)=-14

8. anonymous

I don't think that's how it works though.

9. geerky42

You can just take derivative of each term individually. For each term, use power rule: $$\dfrac{\mathrm d }{\mathrm dx}x^n = nx^{n-1}$$ For constant, you have $$\dfrac{\mathrm d }{\mathrm dx}c = 0$$

10. geerky42

So here, you have $$\dfrac{\mathrm d }{\mathrm dt}(-2-6t) ~~~=~~~ \dfrac{\mathrm d }{\mathrm dt}(-2) + \dfrac{\mathrm d }{\mathrm dt}(-6t) = ~?$$

11. anonymous

(d/-2d)-(d/-2d)=0

12. anonymous

?

13. geerky42

Seem you are not really familiar with basic derivative rules. Can you tell me what you learned recently (what teacher tried to teach you, etc)?

14. anonymous

I don't have a teacher

15. anonymous

Can you tell me what you got, then we can work it out.

16. geerky42

So you are self-taught?

17. anonymous

Probably online school :) ^^

18. anonymous

yes

19. anonymous

If someone could offer me the answer to this question, I would be glad to discuss how it was found