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

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Answer 1 · 2016-10-14T02:44:38

Take the natural logarithm of both sides:

$lny = ln(x^lnx)$

$lny = lnx(lnx)$

$1/y(dy/dx) = 1/x xx lnx + 1/x xx lnx$

$1/y(dy/dx) = lnx/x + lnx/x$

$1/y(dy/dx) = (2lnx)/x$

$dy/dx= ((2lnx)/x)/(1/y)$

$dy/dx= y xx (2lnx)/x$

$dy/dx = x^lnx xx (2lnx)/x$

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

Hopefully this helps!

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