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Integrations and Derivations

  1. Nov 3, 2007 #1
    Ok im in my final study for my last exams... and i just want to check with all you guys that i have my calculus right :)

    1. Lower the power by one, and times by the new power out in front of the term. Coefficinets drop off, and terms with just x become a new coefficient in the f ' (x) form.
    2. e ^ x dervies to:
    (f '(x)) e ^ x
    3. ln (f(x)) dervies to
    [ f '(x)] / [f(x)]
    4. Chain rule
    ( 3x + 4)^5 becomes:
    4(3x + 4)^4 x 3
    =12(3x + 4)^4

    5. Product Rule:
    vu' + uv' where v and u are both functions multiplied together

    6. Quotient Rule:
    (vu' - uv') / (2v)

    f ''(x) is simply the derivative of the derivative

    1. Raise the power by one and mutliply by 1/new power. Add the constant term (c) and any coefficients now add the term x
    eg. x^3 + 2x^2 + 4
    integrates to:
    (1/4)x^4 + (2/3)x^3 + 4x + c

    2. f '(x) / f(x) integrates to ln( f(x)) + c
    3. e ^ f(x) integrates to (1/ f '(x)) e^(f(x)) +c
    4. There are no chain, product or quotient rule to integration
    5. Always add the constant term c, and use a given point to work it out :)

    i hope u guys can identify any probelms in my working if there are any :)
  2. jcsd
  3. Nov 3, 2007 #2
    4 is wrong

    your power is 5, not 4

    6 is wrong

    the denominator is squared not multiplied by 2
  4. Nov 3, 2007 #3
    u mean 4 in derviations?
    doesn't the power always get lowered in deriving things?

    and six. yes my bad thank you :)
  5. Nov 3, 2007 #4
    sorry i wasn't clear enough

    your power was 5, so you bring it down

    you brought down a 4, the reduction process was correct tho
  6. Nov 3, 2007 #5
    ok so.
    4. Chain rule
    ( 3x + 4)^5 becomes:
    4(3x + 4)^4 x 3
    =12(3x + 4)^4

    where the coefficient i brang down is 4 (the new power) is that wrong?
    should it be that i bring down 5 instead (the old power) like so:

    ( 3x + 4)^5 becomes:
    5(3x + 4)^4 x 3
    =15(3x + 4)^4
  7. Nov 3, 2007 #6
    yes, that is correct



  8. Nov 3, 2007 #7
    ok thank you :)

    the ln and e integrations and derviations are ok... they are the ones im not so good at
  9. Nov 3, 2007 #8
    just remember, the derivative of e is itself times the derivative of it's power



    now with a higher power


    [tex]y'=e^{2x^{2}}\times 4x[/tex]

    if your power happens to be a product, then you do the product rule, etc.
  10. Nov 3, 2007 #9
    now for ln



    now a step further


  11. Nov 3, 2007 #10
    okies thats fine :)
    and integration?
  12. Nov 3, 2007 #11
    you have the concept of integration down, but what if your power was a fraction?


  13. Nov 3, 2007 #12
    integration of x^(1/2)
    raise power by one...
    divide by the new power

    = 1/(3/2) x ^(3/2)
    =(2/3) x ^(3/2)
  14. Nov 3, 2007 #13
    yep, it's basically the reciprocal of the power

    you pretty much have it down, just be cool b4 the test. you'll kick it's ass, no worries :-]
  15. Nov 3, 2007 #14
    sweet thanks :) ur help is much appreciated :D
  16. Nov 3, 2007 #15

    btw, you forgot +C ... :p
  17. Nov 3, 2007 #16
    i always forget that :(. but im sure i will remember unless they give me a point on the curve and tell us to work out c, and they generally do that alot cause i have done heaps of past papers.

    just question regarding the graphs of derivatives etc.

    When f '(x) = 0 is a turning point for f(x)
    When f '(x) is positive, the gradient of f(x) is also positive and vice-versa

    when f ''(x) = 0 is a point of inflection for f(x)
    when f "(x) is positive means concave up for f(x) and vice - versa

    is that right?
  18. Nov 20, 2007 #17
    okies final exam done now

    there was much calculus in it and im glad i did my study :)
    thank you i did really well :D
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