How do you find the derivative of $y=x^(ln(x))$ ?

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Answer 1 · 2018-05-04T13:27:42

$d/dx(x^lnx) = 2lnx x^(lnx-1)$

Write the function as:

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

then using the chain rule:

$d/dx(x^lnx) = d/dx (e^(ln^2x)) = e^(ln^2x) d/dx (ln^2x) = (2e^(ln^2x)lnx)/x$

and simplifying:

$d/dx(x^lnx) = 2lnx x^(lnx-1)$

How do you find the derivative of y=x^(ln(x))y=x^(ln(x))?