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[Math] Derivative...

elmoelmo Registered User regular
edited January 2010 in Help / Advice Forum
alright, school asignment so not using real numbers, just need some pointers in the right direction

i have a function
 f(x) = (x^3/600) + 60x + 500 
and need to get the derivative of this, only problem is, i have no idea how to go about the first part, (x^3/600).

Since there is no x under the fraction, i will get 0 there (or will it disapear leaving me with 3x^2?), so i need to some how get rid of the 600 before i can do anything else.

would this be possible?
(x^3/600) + 60x + 500 = 0
(x^3/600) + 60x = -500    | * 600 
 x^3 + 3600x  = -300000

ninja edit: forgot to mention,
i tried moving the 600 above the fraction sign giving me 3.456789e^-4 * x^3, it gives the correct answer with test numbers but looks ugly as sin

or is there some magic/better way of going about this?

elmo on

Posts

  • edited January 2010
    I think your solution is x^2/200+60 but I haven't done derivatives in ages and don't remember the theory of it at all.

    Richard M. Nixon on
    chevy.jpgsteve.jpgmartin.jpg
  • UsagiUsagi Nah Registered User regular
    edited January 2010
    is this x^(3/600) or (x^3)/600?

    Usagi on
  • elmoelmo Registered User regular
    edited January 2010
    Usagi wrote: »
    is this x^(3/600) or (x^3)/600?

    (x^3)/600

    elmo on
  • UsagiUsagi Nah Registered User regular
    edited January 2010
    elmo wrote: »
    Usagi wrote: »
    is this x^(3/600) or (x^3)/600?

    (x^3)/600

    ok

    so remember the derivative of x^3 is 3x^2

    (1/600)*(3x^2) = (1/200)(x^2) = (x^2)/200 or what Nixon said above

    you were massively overthinking this - just remember that constants are always factored out

    Usagi on
  • edited January 2010
    the only reason i remember as much of calc as I do is that we did the painstaking, take a full-page of work derivatives for a week and then my teacher said 'but there's an easier way. put the exponent down in front of x and then reduce the exponent by 1' and he just about had a mutiny on his hands.

    Richard M. Nixon on
    chevy.jpgsteve.jpgmartin.jpg
  • elmoelmo Registered User regular
    edited January 2010
    aaaah, now i get it, brainfart sorted out! was thinking of 600 as a set of its own, not part of the actuall x^3.

    thanks a lot!

    elmo on
  • UsagiUsagi Nah Registered User regular
    edited January 2010
    hey no problem!

    back in the day when I did a lot of calculus, I found it helpful sometimes to think of the equations in terms of Constants x Variables because it made things so much easier to keep separate

    for example, your function f(x) above I would rewrite as C1*x^3 + C2*x + C3, then go through the derivation, and then put the constants back in and evaluate

    eventually I could see the separation in my head and that actually helped tremendously when we moved on to integration

    Usagi on
  • DemerdarDemerdar Registered User regular
    edited January 2010
    Yeah, instead of writing x^3/600 try (1/600)x^3.

    It makes things a lot easier to see.

    Demerdar on
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  • elmoelmo Registered User regular
    edited January 2010
    indeed, i was failing to notice that x^3/600 = (1/600)x^3, other than the brainfart its relativly easy

    elmo on
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