Monday, February 15, 2010

post 27, part 2

ok so i accidentally posted # 27 before i was finished with it. ahah, weird, i know. anyways, here's the rest of post 27.

chain rule:
so it's easier than it seems, right?

we also went over rules for limits approaching 0.
first you plug in, & see what you get. if it ends up somehow with you dividing by zero, (which is 99.9% of the time), then here are the steps you take:
1st try to factor & cancel. if that doesn't work then,
2nd take derivative of top and plug in zero,

exampe: lim sin(x) + 6
x -> 0 x

take derivative of sin(x) = cos(x)
plug in zero, cos(0) = 1. so your answer is 1.

i need help with tan inverse integration. i know she went over it, but i forgot my binder @ school and i don't remember some of it. can someone go over that with me, please? :)

1 comment:

  1. You recognize tan inverse integration by the +1 at the bottom of the fraction.
    To integrate, pull the number out in front then do the tan inverse of the square root of bottom term + c.
    Example: Integrate 9/x^6 + 1
    The +1 hints this to be tan inverse.
    The integral is 9 tan inverse of the square root of x^6 + c
    Therefore, 9 tan inverse of x^3 + c

    Example 2: Integrate 8/ e^2x +1
    8 tan inverse of e^x + c

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