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Math Help - Should be simple differentation

  1. #1
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    Unhappy Should be simple differentation

    But I cant do it

     F(x) = x{a}^{ 1-x} Differentiate with respect to x, so u = x and v = {a}^{ 1-x} But i Cant Differentiate v
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by adam_leeds View Post
    But I cant do it

     F(x) = x{a}^{ 1-x} Differentiate with respect to x, so u = x and v = {a}^{ 1-x} But i Cant Differentiate v
    use algebra

    v=a\cdot a^{-x}=ae^{\ln(x^{-1})}= ae^{-\ln(x)}

    Can you finish from here?
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    use algebra

    v=a\cdot a^{-x}=ae^{\ln(x^{-1})}= ae^{-\ln(x)}

    Can you finish from here?
    \frac{-a}{ x} {e}^{ -ln(x)}

    ?
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    use algebra

    v=a\cdot a^{-x}=ae^{\ln(x^{-1})}= ae^{-\ln(x)}

    Can you finish from here?
    I think there is an algebra error here. I get

    v=a\cdot a^{-x}=a\,e^{\ln(a^{-x})}=a\,e^{-x\ln(a)}}.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    I think there is an algebra error here. I get

    v=a\cdot a^{-x}=a\,e^{\ln(a^{-x})}=a\,e^{-x\ln(a)}}.
    I think too.. I get this same thing...
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    I think there is an algebra error here. I get

    v=a\cdot a^{-x}=a\,e^{\ln(a^{-x})}=a\,e^{-x\ln(a)}}.
    How do you differentiate that wrt to x?
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  7. #7
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    Well, the same way you'd differentiate exp(ax): by using the chain rule. The expression -ln(a) is just a constant, right?
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Well, the same way you'd differentiate exp(ax): by using the chain rule. The expression -ln(a) is just a constant, right?
    Yeah

    so it is -aln(a){e}^{-xln(a) }
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  9. #9
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    The answer for dv should be -{a}^{1-x } lna

    They don't look similar
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  10. #10
    Behold, the power of SARDINES!
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    -a \ln(a)e^{-x\ln(a)}=-a \ln(a)e^{\ln(a^{-x})}= -a\ln(a) \cdot a^{-x}=-a^{1-x}\ln(a)
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