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

1 Answer

Jul 16, 2017

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

Explanation:

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#

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