Question

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

Answer 1

`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`