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L'Hopitals Rule need help with derivatives?

  1. Jul 6, 2006 #1
    Question is:
    Evaluate the following limits
    lim x--> 0 4^x - e^2x / 2x
    So i take derivatives
    but thats where im confused.... Whats the derivate of 4^x... x4^-x???
    and is the derivative of -e^2x -2e^x???
    So then that leaves me with 4x^-x -2e^x /2which is 0... so do i do derivatives again??? It seems like the x's wont go away????
     
    Last edited: Jul 6, 2006
  2. jcsd
  3. Jul 6, 2006 #2

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    You can use the chain rule and the fact that the derivative of e^x is e^x to answer both questions. That is, the derivative of e^f(x) is e^f(x)*f'(x). Try to rewrite 4^x as e^f(x). (use logs)
     
  4. Jul 7, 2006 #3
    [tex]f(x)=4^x[/tex]
    [tex]\ln(f(x))=x\ln(4)[/tex]
    [tex]\frac{1}{f(x)}*f'(x)=\ln(4)[/tex]

    You can take it from there.
     
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