How do you differentiate $f(x)=ln(x)^x$ ?

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Answer 1 · 2016-01-20T01:04:36

$f'(x) = ln^x(x)(1/ln(x)+ln(ln(x)))$

Let $y = ln^x(x)$

$=>ln(y) = ln(ln^x(x)) = xln(ln(x))$

$=>d/dxln(y) = d/dxxln(ln(x))$

$=>1/ydy/dx = x(d/dxln(ln(x))) + ln(ln(x))(d/dxx)$

$=x(1/(ln(x))(d/dxln(x))) + ln(ln(x))$

$=x(1/ln(x)(1/x)) + ln(ln(x))$

$= 1/ln(x) + ln(ln(x))$

$=> dy/dx = y(1/ln(x) + ln(ln(x)))$

$= ln^x(x)(1/ln(x) + ln(ln(x)))$

How do you differentiate f(x)=ln(x)^xf(x)=ln(x)^x?