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Higher Order Derivatives

  1. Oct 31, 2005 #1
    Hello Everyone,

    I'm doing some questions on higher derivatives, and they should be easy but I am worried that my answers are not quite right. There just seems to be something 'off' about them. Anyway here are the questions and my answers.

    1) Find the first and second derivatives of y= (4x)/squareroot(x+1)

    y' =(squareroot(x+1))(4)-(4x)(1/2)((x+1)^-1/2))/(x+1)


    Ok, so that was my first derivative and I think it is ok, now for the second:

    y''=((x+1)[(2(x+1)^-1/2)-(2(x+1)^-1/2)-(x(x+1)^-3/2)] -4squareroot(x+1)+2x(x+1)^-1/2)/(x+1)^2

    I hope that doesn't look to confusing but I have a bad feeling it is.

    2) Find the first and second derivatives of x^4+y^4=a^4

    My biggest issue with this one is that I'm unsure of how to deal with the a, I just treated it like the x, and used implicit differentiation:

    x^4 + y^4 = a^4

    4x^3 + 4y^3 y' = 4a^3

    y' =(4a^3 - 4x^3)/4y^3

    y' = (a^3 -x^2)/y^3

    y'' = y^3(3a^2-3x^2)-3y^2(a^3-x^3)/(y^6)

    y'' = 3y^2 (ya^2-yx^2-a^3+x^3)/y^6

    y'' = (3(ya^2-yx^2-a^3+x^3))/y^4

    Well there it is, hope what I've shown makes sense, I know it is hard to read on a computer. Thanks for any help guys, I really appreciate your time and effort.
  2. jcsd
  3. Oct 31, 2005 #2
    in the second problem 1 is a constant ant dy/dx of a constant is zero
  4. Oct 31, 2005 #3


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    Sometimes, differentiating quotients, it is simpler to use a negative exponent and use the product rule rather than the quotient rule.

    For (1) write the function as y= 4x(x+1)1/2.
    Then y'= 4(x+1)1/2+ (4x)(1/2)(x+1)-1/2.
    = 4(x+1)1/2+ 2x(x+1)-1/2
    Now y"= 4(1/2)(x+1)-1/2+ 2(x+1)-1/2-(2x)(1/2)(x+1)-3/2
    = 4(x+1)-1/2- x(x+1)-3/2.
    Since the original problem was written as a fraction, you might want to write that as (3x+ 4)/(x+1)3/2 or

    As mathmike pointed out, the "a^4" is a constant. Its derivative is 0. You have x4+ y4= a4 so 4x3+ 4y3y'= 0 or simply x3+ y3y'= 0. Now it's better not to solve for y' but just use implicit differentiation again:
    3x2+ 3y2y'+ y3y"= 0
    Now [tex]y"= -\frac{3x^2+ 3y^2y'}{y^3}[/tex].
    That still has y' in it but that's okay.
  5. Oct 31, 2005 #4


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    Uhmm,... there's a slight error in it.
    It should read: y= 4x(x+1)-1/2, not y= 4x(x+1)1/2 :smile:
    y'= 4(x+1)-1/2 - (4x)(1/2)(x+1)-3/2.
    = 4(x+1)-1/2 - 2x(x+1)-3/2 = (4x + 4 - 2x)(x+1)-3/2 = (2x + 4)(x+1)-3/2.
    y'' = 2(x+1)-3/2 - 3 / 2 (2x + 4)(x+1)-5/2
    = (2x + 2 - 3x - 6)(x+1)-5/2 = (-x - 4)(x+1)-5/2
    [tex]= -\frac{x + 4}{\sqrt{(x + 1) ^ 5}}[/tex].
    Viet Dao,
  6. Oct 31, 2005 #5
    Hmmm ok, I never realized that a should be taken as a constant. Now that I know you have to take it as a constant, could someone please explain why to me? How do you know a is a constant instead of just another variable, I had a question awhile ago where we had to take the derivatives of d's and b's and a's all together in a question and I'm not sure how to tell when you take it as a constant or not. As for the other questions thanks again for the help, I will go and try it using the product rule as you suggested and see if I can get it to work out that way. Thanks a lot.
  7. Oct 31, 2005 #6
    the derivative of x^n = nx^(n-1)

    and if it is ax^n it would be anx^(n-1)

    and d/dx[a]=0

    so if you take the der of something that is multiplied by a constant then the der is still multiplied by the constant.

    but if it has no variable then it will go to zero
  8. Oct 31, 2005 #7
    OK thanks again. I understand that the derivative of a constant is zero, but I guess what I am trying to ask is how to you know for sure that a is a constant? I am asking this because I have had a question before where you had to differentiate a's and b's and there were no x's in front of them, you did it as though a was not a constant. Maybe I am just being stupid here...haha.
  9. Oct 31, 2005 #8


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    The problem asked you to find dy/dx. The only way you can do that is if y is a function of the single variable x: a can't be a variable.

    You probably haven't covered partial derivatives yet, but even if you were given that y was a function of the two variables x and a, the partial derivative [tex]\frac{\partial y}{\partial x}[/tex] would be done by treating a as if it were a constant so you would get the same answer.
  10. Oct 31, 2005 #9
    Oh, ok I think I understand now, thank you!
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