First and Second Derivatives

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Just trying to find the first and second derivatives.

    X^2/(X^2-16)

    1+X/1-X

    X^3(X-2)^2

    2. Relevant equations
    Quotient Rule/Power Rule/Chain Rule


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 6, 2012 #2
    For the first one, what is the derivative of f(x)/g(x)? In general, not for this specific function.
     
  4. Mar 6, 2012 #3
    (g(x)*f'(x)-f(x)*g'(x))/g(x)^2
     
  5. Mar 6, 2012 #4
    okay, so if f(x) = x^2 and g(x) = x^-2, what is the derivative of f(x)/g(x)?
     
  6. Mar 6, 2012 #5
    f(x) = x^2 and g(x) = x^-2
    x^2/x^-2
    ((x^-2)(2x)-(-2x)(x^2))/ (x^2)^2
     
  7. Mar 6, 2012 #6
    I mistyped (that g(x) should have been g(x)=x^2 - 16 [not sure how I messed that one up]), but you seem to know what you're doing. What specific question do you have?
     
  8. Mar 6, 2012 #7
    I'm mostly having trouble with the second derivatives.
     
  9. Mar 6, 2012 #8
    What do you have so far? Where are you getting stuck?
     
  10. Mar 8, 2012 #9
    Well here's what I've got, I think they're right but I'm not sure.

    f(x)=X^2/(x^2-16)
    (X^2-16)(2X)-(X^2)(2X)/(X^2-16)^2
    f'(x)=-32X^2/(X^2-16)^2
    (X^4-32X^2+256)(-64X)-(-32X^2)(4X^3-64X)/(X^4-32X^2+256)^2
    -64x^5+2048X^3-16384X+128X^5-2048X^3
    f''(x)=64X^5-16384X/(X^4-32X^2+256)^2

    f(x)=1+x/1-X
    (1-X)(1)-(1+X)(-1)/(1-X)^2
    f'(x)=2-2X/(1-X)^2
    (X^2-2X+1)(-2)-(2X-2)(2-2X)/(X^2-2X+1)^2
    f''(x)=2X^2+4X+2/(X^2-2X+1)^2

    f(x)=X^3(X-2)^2

    (X^3)(2X-4)+(3X^2)(X-2)^2
    2X^4-4X^3+3X^4-12X^3+12X^2
    f'(x)=5X^4-16X^3+12X^2
    5X^4-16^3+12X^2
    f''(x)=20X^3-48X^2+24X
     
  11. Mar 8, 2012 #10
    Am I doing these correctly?
     
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