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Derivatives of constants

  1. Sep 18, 2005 #1
    this is just a general question, one that keeps coming up everytime i take a test or quiz.

    I understand that the derivative of a constant is zero, but I dont understand when you use that zero to multiply or divide in a problem, and when you just ignore that zero.

    For example x^2/3
    Would that be zero? or 2x?
    Because I know in some problems you just ignore the constant when it is alone, but I'm not sure about when it is a variable.

    Help is appreciated, I have a test tomorrow
    Thanks
     
  2. jcsd
  3. Sep 18, 2005 #2

    TD

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    The derivative of a constant term is 0, in your case the term isn't constant. The "3" you're probably referring to is a factor which belongs to the term "x²/3".

    Since the derivative is lineair, you can put the factor up front and get:

    [tex]\left( {\frac{{x^2 }}
    {3}} \right)^\prime = \frac{1}
    {3}\left( {x^2 } \right)^\prime = \frac{{2x}}
    {3}[/tex]
     
  4. Sep 18, 2005 #3

    Doc Al

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    If y = a [a constant by itself], then [itex]dy/dx = 0[/itex].
    If y = a f(x) [a constant times a function], then [itex]dy/dx = a (df/dx)[/itex]

    So if [itex]y = (1/3)x^2[/itex], then [itex]dy/dx = (1/3)(2x) = (2/3)x[/itex]
     
  5. Sep 18, 2005 #4

    Hurkyl

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    Try applying the division rule to this. (of course, it's better if you remember the scalar multiplication rule, but since this is a sticky point for you, try doing it with the division rule and see if you're happy)
     
  6. Sep 18, 2005 #5

    HallsofIvy

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    I don't think that is his problem, Hurkyl!

    Because (f+ g)'= f'+ g', (f+ Constant)'= f'+ Constant'= f'+ 0= f' so you can "ignore" the constant. (Notice the quotes- you are not IGNORING it! You are THINKING about it, recognizing that its derivative is a constant and so "adding 0".

    Because (fg)'= f'g+ fg', (Constant*g)'= (Constant)'g+ (Constant)g'= 0*g+ Constant*g'. The derivative of a constant times a function is the constant times the derivative of the function. I guess you could say you are "ignoring" the constant here but that's why I don't like that word! THINK about what you are doing!

    Surely you know that every function involves some constants (even if they are 1). You can't just "ignore" them.

    The two rules I mentioned above: (f+ Constant)'= f' and (Constant*f)'= Constant*f' are as close as you can come to "ignoring" the Constant!
     
  7. Sep 18, 2005 #6

    Hurkyl

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    I'm hoping he'll appreciate what's going on better if he does the division rule and sees the constant resurface at the end. :smile: In effect, I'm hoping he'll work out for himself exactly what you told him about the product rule.
     
  8. Sep 18, 2005 #7
    i'm not seeing a constant there. a quick mental calculation gives [tex]\frac{d}{dx}f(x)=\frac{2}{3}x^{-\frac{1}{3}}[/tex]
     
  9. Sep 18, 2005 #8

    Doc Al

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    It depends on what the OP meant by "x^2/3": either (x^2)/3 or x^(2/3). (From the context I assumed he meant the former.)
     
  10. Sep 18, 2005 #9
    that's why you use [tex]\LaTeX[/tex]
     
  11. Sep 18, 2005 #10

    Hurkyl

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    And, of course, the strict interpretation x^2/3 is (x^2)/3.
     
  12. Sep 18, 2005 #11

    Doc Al

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    And that's even true in Latex: x^2/3 = [itex]x^2/3[/itex]. :smile:
     
  13. Sep 18, 2005 #12
    then it's [tex]\frac{d}{dx}f(x)=\frac{2}{3}x[/tex]
     
  14. Sep 19, 2005 #13

    HallsofIvy

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    I was thinking that hotrocks007 meant x^(2/3) but was thinking "Aha- there's a constant, 2/3, in there and the derivative of a constant is 0! So the derivative of this is 0!" That was the reason for my remark that every function contains "constants"!
     
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