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Thread: Problem:Calculus: Find derivative of [x (x+1) (x+2) (x+3)]^(1/3)

  1. #1
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    Problem:Calculus: Find derivative of [x (x+1) (x+2) (x+3)]^(1/3)

    "I'm a college freshman in calculus, and we;re studying derivatives."

    Q.Find the derivative of the following function:-
    1)3th-rt(x(x+1)(x+2)(x+3)) .... this can be written as [x(x+1)(x+2)(x+3)](1/3)

    My work:-

    Webp.net-resizeimage.jpg

    Webp.net-resizeimage (1).jpg

    The correct answer is:- -4/(e^2x-1)

    while i am stuck at a point where i have two options either 1)(x+2)(3x+1) 2)x(x+1)(x+1)(x+2) and none of it seems to be correct.
    So,i need help in this part.
    Last edited by Subhotosh Khan; 10-06-2017 at 08:04 AM.

  2. #2
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    Quote Originally Posted by manishpamnani169 View Post
    "I'm a college freshman in calculus, and we;re studying derivatives."

    Q.Find the derivative of the following function:-
    1)3th-rt(x(x+1)(x+2)(x+3)) .... this can be written as [x(x+1)(x+2)(x+3)](1/3)

    My work:-

    Webp.net-resizeimage.jpg

    Webp.net-resizeimage (1).jpg

    The correct answer is:- -4/(e^2x-1)

    while i am stuck at a point where i have two options either 1)(x+2)(3x+1) 2)x(x+1)(x+1)(x+2) and none of it seems to be correct.
    So,i need help in this part.
    If I were to do this problem I would start it this way:

    d/dx{ [x(x+1)(x+2)(x+3)](1/3)}

    =d/dx{ [x(1/3)] * [(x+1)(x+2)(x+3)](1/3)}

    = [x(1/3)] * d/dx{[(x+1)(x+2)(x+3)](1/3)} + (1/3) * [x(-2/3)] * [(x+1)(x+2)(x+3)](1/3)

    = [x(1/3)] * d/dx{[(x+1)(1/3)][(x+2)(x+3)](1/3)} + (1/3) * [x(-2/3)] * [(x+1)(x+2)(x+3)](1/3)

    = [x(1/3)] * {[(x+1)(1/3)]d/dx{[(x+2)(x+3)](1/3) + (1/3) * [(x+1)(-2/3)]*[(x+2)(x+3)]} + (1/3) * [x(-2/3)] * [(x+1)(x+2)(x+3)](1/3)

    and continue .... (be careful about those parentheses)
    Last edited by Subhotosh Khan; 10-06-2017 at 08:25 AM.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Thanks!!

    Quote Originally Posted by Subhotosh Khan View Post
    If I were to do this problem I would start it this way:

    d/dx{ [x(x+1)(x+2)(x+3)](1/3)}

    =d/dx{ [x(1/3)] * [(x+1)(x+2)(x+3)](1/3)}

    = [x(1/3)] * d/dx{[(x+1)(x+2)(x+3)](1/3)} + (1/3) * [x(-2/3)] * [(x+1)(x+2)(x+3)](1/3)

    = [x(1/3)] * d/dx{[(x+1)(1/3)][(x+2)(x+3)](1/3)} + (1/3) * [x(-2/3)] * [(x+1)(x+2)(x+3)](1/3)

    = [x(1/3)] * {[(x+1)(1/3)]d/dx{[(x+2)(x+3)](1/3) + (1/3) * [(x+1)(-2/3)]*[(x+2)(x+3)]} + (1/3) * [x(-2/3)] * [(x+1)(x+2)(x+3)](1/3)

    and continue .... (be careful about those parentheses)
    Thanks for the reply i will definitely try it out!!

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