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Derivatives! Please, help me, I have an exam tomorrow!

  1. Feb 21, 2013 #1
    [itex]f(x)=5x^3+6x^2-3x+lnx[/itex]
    (lnx)`=1/x

    [itex]f(x)=2x^4+3x^2+ cosx[/itex]
    (cosx)`=-sinx


    I know that if I only have x, like 3x, then x disappears (correct me if I'm wrong). So what happens with lnx if x disappears?
    Same thing with cosx.

    The lesson is extreme values of functions and i saw critical points mentioned a lot and Fermat's theorem, and Sylvester's, but I was told I should understand derivatives first.

    I have spent the last day trying to understand these, from manuals and internet and I can't find anything clear and simple enough.
    I used to study arts and psychology and I somehow ended in economics, Please help!!! and please explain as if you are talking to a 6th grade child so I can understand.
    Or if you can direct me to something where these things are really, really dumbed down so I can understand it. I'd be very grateful.
     
    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 21, 2013 #2
    I think you're confused on the power rule. This is how it is typically written:
    [tex]\frac{d(x^{n})}{dx}=nx^{n-1}[/tex]
    So what this is saying is if you have a function like x^2, what you do is take the n (2 in this case), bring it infront of the x and subtract 1 from the exponent. Then it becomes 2x^1=2x.

    So for your first example, 5x^3 will be 15x^2 (and so on)

    As for the lnx part, you don't do anything to the x. The derivative of lnx is simply just 1/x.
     
  4. Feb 21, 2013 #3
    Thank you iRaid, that is exactly what i needed to know.
    Next, I dont understand this:
    if I have
    f(x,y)=3xy-x^3-y^3
    first I want to know how do you solve f(x)= then f(y)=
    then how do you find out the critical points.
    I'm sorry if this is stupid, but I have the solved exercise in front of me and I still can't figure out what and why is happening.

    And with cosx, should I know it's value and replace it or do I write
    f`(x)=8x^3+6x+cosx?
     
  5. Feb 21, 2013 #4

    HallsofIvy

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    By the way, on your test (or anywhere else but especially on a test!) don't write things like "ln x= 1/x" and "sin x= cos x". You mean "the derivative of ln x is 1/x" or "(ln x)'= 1/x" and "the derivative of sin x is cos x" or "(sin x)'= cos x". NEVER write "=" between things that are NOT equal.
     
  6. Feb 21, 2013 #5
    Thank you for correcting me, it's how I wrote it in class when the teacher was speaking, I had no idea they are not equal.
     
  7. Feb 21, 2013 #6

    HallsofIvy

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    I'm sorry, I thought it was just an oversight. Are you really saying that you thought ln(x)= 1/x and sin(x)= cos(x)??
     
  8. Feb 21, 2013 #7
    Yes, last time I studied math before this was years ago, in highschool, and I forgot all of it. I even had to repeat basic things like positive and negative numbers and exponents and fractions.
     
  9. Feb 21, 2013 #8
    If your teacher wrote ln(x)=1/x then he made a pretty big mistake. He should of put d(ln(x)/dx=1/x because I can see why that would lead to major confusion (especially if you haven't done math in a while).

    I'm not sure by what you mean:
    Do you mean solve f'(x) and f'(y)?
     
  10. Feb 21, 2013 #9
    Yes, solve. And find out the critical points.
    And if you could explian what a critical point is, that would be great too.
    I read the wikipedia article about critical points and didn't understand a thing.
     
  11. Feb 21, 2013 #10
    Well you would have to do implicit differentiation for the problem above.
    Critical points are the relative/absolute minimum/maximum. To find the relative minmum/maximum you take the second derivative (I think its second, don't remember exactly) of the function and set it equal to 0.
     
  12. Feb 22, 2013 #11
    Do you know how to calculate limits?
    Do you know what's the definition of a derivative?
    If you know I can explain to you how to get the derivatives of some functions, if you don't you'll have to begin from limits to really understand this.
     
  13. Feb 22, 2013 #12
    The derivative of a function is a certain attribute of that function.

    The derivative of 3x is 3, because the function 3x is always changing at a rate of 3.

    But obviously, the function 3x is not the same as the number 3.

    As for your leading question, in 3x, the "rule" is not that x disappears, the rule is the power rule. However, the result is that x disappears in this case.

    Since 3x = 3x^1, I take the derivative by multiplying 3 by the exponent 1, and lowering the exponent by 1.

    3(1)x^(1-1) = 3(1)x^0 = 3(1)(1) = 3

    And that is really why x disappears.
     
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