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Math Help - Tricky Logarithmic Differentiation

  1. #1
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    Tricky Logarithmic Differentiation

    y=lnx^(tanx) or y= lnx^(tanx).

    So: with logarithmic differentiation:

    lny= ln(lnx^(tanx))

    1/y dy/dx = tanx ln (lnx)

    1/y dy/dx = sec²x ln(lnx) + tanx(ln(1/x)
    1/y dy/dx= sec²xln(lnx) + tanx(???)

    My Question Is: What would be the derivative of ln(1/x)

    The final answer is: dy/dx= (lnx)^tanx * sec²ln(lnx) + tanx/(xlnx)

    How is the derivative of ln(1/x)= xlnx?

    Thank you.
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  2. #2
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    Quote Originally Posted by r2d2 View Post
    y=lnx^(tanx) or y= lnx^(tanx).

    So: with logarithmic differentiation:

    lny= ln(lnx^(tanx))

    1/y dy/dx = tanx ln (lnx)

    1/y dy/dx = sec²x ln(lnx) + tanx(ln(1/x)
    1/y dy/dx= sec²xln(lnx) + tanx(???)

    My Question Is: What would be the derivative of ln(1/x)

    The final answer is: dy/dx= (lnx)^tanx * sec²ln(lnx) + tanx/(xlnx)

    How is the derivative of ln(1/x)= xlnx?

    It isn't: \ln\frac{1}{x}=-\ln x\Longrightarrow \left(\ln\frac{1}{x}\right)'=-\frac{1}{x}

    Tonio

    Thank you.
    .
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