How do you find the derivative of $cosx^tanx$ ?

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Answer 1 · 2017-07-16T15:12:33

$d/dx(cosx^tanx)=(sec^2xlncosx-tan^2x)cosx^tanx$

First define $y= cosx^tanx$

Then, by definition, $lny =tanxlncosx$

And $1/y(dy/dx) = sec^2 xlncosx -(sinx/cosx)tanx$

$dy/dx=y(sec^2xlncosx - tan^2x)$
$=(sec^2xlncosx-tan^2x)cosx^tanx$

How do you find the derivative of cosx^tanxcosx^tanx?