How do you find the derivative of $(cosx)^x$ ?

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Answer 1 · 2015-10-05T20:20:57

$y'=(cosx)^x(ln(cosx)-xtgx)$

$y=(cosx)^x$
$lny=ln(cosx)^x$
$lny=xln(cosx)$

$d/dx(lny)=d/dx(xln(cosx))$

$1/y*y'=ln(cosx)+x*1/cosx*(-sinx)$

$(y')/y=ln(cosx)-xtgx$

$y'=y*(ln(cosx)-xtgx)$

$y'=(cosx)^x(ln(cosx)-xtgx)$

How do you find the derivative of (cosx)^x(cosx)^x?