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Differentiation of exponential

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    y = x^(18x^-4)

    2. Relevant equations

    chain rule
    dy/dx a^x = a^x ln a

    3. The attempt at a solution
    first i used the second equation from above to get
    x^(18/x^4)*ln(x)
    then i use the chain rule to get
    x^(18/x^4)*ln(x)*(-72x^-5)
    the computer program i am using is saying this is wrong and the only thing i can think of that is wrong is the fact I may not be able to use the one formula since x is not a constant but if this is true then i have no idea what to do
     
  2. jcsd
  3. Oct 22, 2009 #2
    Try taking the log of both sides and using implicit differentiation.
     
  4. Oct 22, 2009 #3
    alright so then i should get
    ln y = ln (x^(18/x^4))
    1/y = 18/x^4 ln x
    = y 18/x^4*1/x
    = y 18/x^5
    = x^(18/x^4)*18/x^5
    ?
     
  5. Oct 22, 2009 #4
    well the computer told me that solution is wrong also so any other advice would be helpful
     
  6. Oct 22, 2009 #5

    Office_Shredder

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    This is only true if a is a constant! Do you know how to differentiate xx? Hint: Try writing it as ef(x) for some function f(x). The same technique can be applied here.

    It doesn't look like you took the derivative of the right hand side in your other post correctly
     
  7. Oct 22, 2009 #6
    does this look right
    ln y= 18/x^4 ln x
    1/y= 18/x^5 + -72x^-5 ln x
    = y 18/x^5 + -72x^-5 ln x
    then just plug in y ?
     
  8. Oct 23, 2009 #7
    yep!
     
  9. Oct 23, 2009 #8

    HallsofIvy

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    You've forgotten the "dy/dx"!

    and what is this equal to?

    If [itex]ln y= (18/x^4)ln(x)[/itex]
    then [iitex](1/y)(dy/dx)= (18/x^5) -72x^{-5}ln x[/itex]
    Now solve for dy/dx by multiplying both sides by y.
    You might not be required to substitute for y. It depends on exactly what is wanted.
     
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