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Math Help - Chain rule - what am i doing wrong?

  1. #1
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    Chain rule - what am i doing wrong?

    I want to take the derivative of  \left(\frac{1976}{x}\right)^x

    Using the chain rule, i got  \left(\frac{-1976}{x}\right)\left(\frac{1976}{x}\right)^{x-1}
    But I know this is an increasing function (so the derivative should not be negative) and so I am kind of confused. Please help.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by machack View Post
    I want to take the derivative of  \left(\frac{1976}{x}\right)^x

    Using the chain rule, i got  \left(\frac{-1976}{x}\right)\left(\frac{1976}{x}\right)^{x-1}
    But I know this is an increasing function (so the derivative should not be negative) and so I am kind of confused. Please help.
    Not as easy as it seems. Check here. Click on Show Steps.
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by machack View Post
    I want to take the derivative of  \left(\frac{1976}{x}\right)^x

    Using the chain rule, i got  \left(\frac{-1976}{x}\right)\left(\frac{1976}{x}\right)^{x-1}
    But I know this is an increasing function (so the derivative should not be negative) and so I am kind of confused. Please help.
    hmmm..

    let y = (\frac{1976}{x})^x

     lny = xln( \frac{1976}{x} ) = xln(1976) -xlnx

    Implicitly differentiate

     \frac{1}{y} y^{ \prime} = ln(1976) - (lnx + 1)

     y^{ \prime } = y[ln(1976) - (lnx + 1) ]

     y^{ \prime } = (\frac{1976}{x})^x [ln(1976) - (lnx + 1) ]
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  4. #4
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    Ah I just realized what i did wrong (thought a^x could be differentiated into x*a^(x-1))
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