Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

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#

Related questions

See all questions in Derivative Rules for y=cos(x) and y=tan(x)

Impact of this question

1177 views around the world

You can reuse this answer
Creative Commons License