Calculus help



  • What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
    nx^(x-1), but couldn’t see how this was implemented.

    If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.



  • d/dx(x^n) = n x^n-1 is the right rule to use.

    d/dx (x^1/3) = (1/3) * x^(1/3 - 1) = (1/3) *x^(-2/3)

    d/dx (x^-1/3) = (-1/3) *  x^(-1/3 - 1) = (-1/3) *x^(-4/3)

    You can not eliminate the 0 from the denominator
    The two derivative function are both defined in all R (set of the real numbers) excluding 0, where they are not defined and 0 is a point of discontinuity.


  • 2007 AAR League

    Lmao!
    I’m about to start posting my homework on this site.


  • 2018 2017 '16 '11 Moderator

    Can I post my CompSci homework here?

    Anyone know C++???



  • @Romulus:

    d/dx(x^n) = n x^n-1 is the right rule to use.

    d/dx (x^1/3) = (1/3) * x^(1/3 - 1) = (1/3) *x^(-2/3)

    d/dx (x^-1/3) = (-1/3) *  x^(-1/3 - 1) = (-1/3) *x^(-4/3)

    You can not eliminate the 0 from the denominator
    The two derivative function are both defined in all R (set of the real numbers) excluding 0, where they are not defined and 0 is a point of discontinuity.

    Thanks for the help.

    When referring to removing 0 from the denominator, you were talking about the two derivative functions you solved for me, no?



  • @Nukchebi0:

    What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
    nx^(x-1), but couldn’t see how this was implemented.

    If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

    You said that you tried to eliminate the 0 from the denominator.
    I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^(1/3) is continuous and defined in all the set of real number R. Its derivated function (1/3) * x^(-2/3) is discontinuous in the point 0, where it is not defined.
    If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.



  • @Commander:

    Can I post my CompSci homework here?

    Anyone know C++???

    I am an ex C/C++ programmer. After I passed to Java and nowe I am pratically an analyst/designer so I do not write code anymore!  😢
    And this is areally bad thing for me!

    I do not remember all the C++ syntax and rules, but you may try to ask!  🙂



  • @Romulus:

    @Nukchebi0:

    What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
    nx^(x-1), but couldn’t see how this was implemented.

    If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

    You said that you tried to eliminate the 0 from the denominator.
    I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
    If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.

    I see now, thanks.



  • @Nukchebi0:

    @Romulus:

    @Nukchebi0:

    What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
    nx^(x-1), but couldn’t see how this was implemented.

    If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

    You said that you tried to eliminate the 0 from the denominator.
    I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
    If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.

    I see now, thanks.

    Excuse me, in the post you have quoted I wrote the derivate worng it is : (1/3)*x^(-2/3)!
    But the reasoning is right.
    I have corrected the original post!



  • boy - i hope that Romulus gets a lot of positive Karma for this . . . .



  • @Romulus:

    @Nukchebi0:

    @Romulus:

    @Nukchebi0:

    What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
    nx^(x-1), but couldn’t see how this was implemented.

    If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.

    You said that you tried to eliminate the 0 from the denominator.
    I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
    If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.

    I see now, thanks.

    Excuse me, in the post you have quoted I wrote the derivate worng it is : (1/3)*x^(-2/3)!
    But the reasoning is right.
    I have corrected the original post!

    I figured that you did. I make a ton of mistakes of that ilk.

    @cystic:

    boy - i hope that Romulus gets a lot of positive Karma for this . . . .

    And yes, he will receive positive karma.

    Edit: It seems I can’t. I try and I get a message saying server verification has failed. The suggested fix does nothing.


  • 2018 2017 '16 '11 Moderator

    Prolly not a big deal.  I just really, really, REALLY hate coding.



  • I love coding! But in the last year I am coding less and less.
    I am ususally involved in analysis and design.
    I have to write a lot of documentation!

    I am happy that at least I am writing the docs in Latex so I still have to “compile” something!



  • my brain hurts.



  • @cyan:

    my brain hurts.

    Sorry!  :oops:

    Next time I will use PM!  😄
    All in all this is an A&A forum…


  • 2007 AAR League

    @cystic:

    boy - i hope that Romulus gets a lot of positive Karma for this . . . .

    He should get negative Karma … how will nukchebi0 learn now?!



  • @AJ:

    @cystic:

    boy - i hope that Romulus gets a lot of positive Karma for this . . . .

    He should get negative Karma … how will nukchebi0 learn now?!

    I learnt when someone showed to me how to apply derivative rules.
    If you try to apply a mathematical rule and it do not work it pointless to continue to try. Mathematics is a correct science, someonw have to show you how to apply it correectly.

    Do, or do not. There is no try.


  • 2018 2017 '16 '11 Moderator

    There is no wrong answer in Mathematics.  Just a procedure that didn’t work the way you had hoped.  Sometimes a U-substitution works, sometimes Integration by Parts, sometimes FOIL, sometimes LOGs/Exponentials, etc.



  • Is the thing I tried to say…
    I agree with you!


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