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I can't do derivative of:(1025t^2-800t+400)^(1/2)

  1. Jan 30, 2005 #1
    I can't do derivative of:

    (1025t^2-800t+400)^(1/2)
     
  2. jcsd
  3. Jan 30, 2005 #2

    dextercioby

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    Do you know the chain rule??If so,apply it.

    Daniel.
     
  4. Jan 30, 2005 #3
    use chain rule.....
    the derivative of x^1/2 = 1/2 x^-1/2

    chain rule and above formulas should solve your problem... if you have no idea how to apply chain rule, look it up in your textbook and study carefully, you must fully understand before you goto the next chapter, otherwise, you will have a tough time in the future.
     
  5. Jan 30, 2005 #4
    thanks you :)

    ya im doing Optimization in Economics and Science......SO DAMN HARD

    so I end up getting f1(x)=(1/2)(1025t^2-800t+400)^(-1/2)(2050t-800)

    THen my teacher jumps too: If f1(x)=0 2050t=800

    WTF?
     
    Last edited: Jan 30, 2005
  6. Jan 30, 2005 #5

    Zurtex

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    I asusme you want to differentiate with respect to t.

    Let:

    [tex]u=1025t^2-800t+400[/tex]

    Then your function is:

    [tex]u^{\frac{1}{2}}[/tex]

    By the chain rule:

    [tex]\frac{d \left( u^{\frac{1}{2}} \right)}{dt} = \frac{1}{2} \frac{du}{dt} u^{- \frac{1}{2}}[/tex]

    You should be able to take it from there.
     
    Last edited: Jan 30, 2005
  7. Jan 30, 2005 #6
    thanks you :)

    ya im doing Optimization in Economics and Science......SO DAMN HARD

    so I end up getting f1(x)=(1/2)(1025t^2-800t+400)^(-1/2)(2050t-800)

    THen my teacher jumps too: If f1(x)=0 2050t=800

    WTF?
     
  8. Jan 30, 2005 #7
    He/she is finding turning points?
     
  9. Jan 30, 2005 #8
    trying to find t, sets 0 to find max or min
     
  10. Jan 30, 2005 #9

    dextercioby

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    Not really,u should compute the second derivative as well.It's the only way you can decide which is max and which is min,if any...

    Daniel.
     
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