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March 25, 2017

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find the der of

y= x^(lnx)

thanks

  • Calc--derivatives - ,

    Use the function of a function rule, namely
    dy/dx = dy/dy . du/dx
    substitute u=log(x)
    and find the derivative of
    d(x^u)/du then multiply by du/dx.

  • Calc--derivatives - ,

    ummm.. i don't understand your language!

    can you explain that in baby terms? thanks!

  • Calc--derivatives - ,

    how about this
    take ln of both sides

    ln y = ln(x^lnx)
    ln y = lnx(lnx) = (lnx)^2

    now differentiate implicitly
    (dy/dx)/y = 2lnx * 1/x
    dy/dx = y(2lnx)/x
    = (x^lnx) (2lnx)/x

    somebody please check this, I have been making some silly mistakes today.

  • Calc--derivatives - ,

    did u get

    e^(lnx) / x^2 ??

  • Calc--derivatives - ,

    I will try with an example.

    Use the function of a function rule, namely
    dy/dx = dy/dy . du/dx
    substitute u=log(x)
    and find the derivative of
    d(x^u)/du then multiply by du/dx.

    Suppose y=sin(x²)
    We know how to differentiate sin(x), and how to do x², but we're not sure how to find the derivative of sin(x²).

    So let u=x²
    then
    y=sin(u)
    dy/du = cos(u)
    We also know that
    du/dx = d(x²)/dx = 2x

    So, by the chain rule,
    dy/dx = dy/du.du/dx = 2xcos(x²)

    You can apply the same procedure to your problem using u=ln(x).

  • Calc--derivatives - ,

    For Carolina,
    There is no e^x in the answer.
    The derivative of log(x) is 1/x.

    Also,
    Reiny's answer is right.
    Mine comes up to
    2*x^(log(x)-1)*log(x)
    which is the same thing.

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