once again

Okay, Review of old stuff..its what I do!!


Complex Derivatives y=ln(e^x) (Chain Rule)

First off one should should identify the steps of your problem. In this case they would be:

1. Natural Log
2. e^x

you problem should be (1/(e^x)).(e^x)'

then you find the derivative of e^x which is e^x . x' (x'=1)

so your final problem should be (1/(e^x)).(e^x)

After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.


First Derivative Test:

1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)

Second Derivative Test:

1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)


UM....okay.. I need help on velocity and position problems. I get the whole concept..and general thing. but some of those problems on those AP's..yeah..I see a keyword and skip it...so if you could help me...much appreciated.

and keeping with chelsea's comment, do we have to still do comments and just send them to brob..or what???? Well, have a good holidays.