How do you differentiate $y=x^(x-1)$ ?

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Answer 1 · 2016-12-21T12:28:07

$dy/dx= x^(x-1)*((x-1)/x+lnx)$

$y=x^(x-1)$

$lny = (x-1)*lnx$

$1/ydy/dx= (x-1)*1/x+lnx*1$

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

$dy/dx= x^(x-1)*((x-1)/x+lnx)$

How do you differentiate y=x^(x-1)y=x^(x-1)?