Question

How do I find the derivative of `y=tanx^secx + secx^cotx`?

Answer 1

`dy/dx = (secxtanxlntanx+sec^2cscx)tanx^secx + (1-csc^2xlnsecx)secx^cotx`

Explanation:

Let `y= tanx^secx + secx^cotx`

Let `u=tanx^secx` and `v=secx^cotx`

`lnu = secxlntanx`

`1/u` `(u')= secxtanxlntanx + sec^2xcscx`

`u' = u(secxtanxlntanx + sec^2xcscx) = (secxtanxlntanx+sec^2xcscx)tanx^secx`

`lnv = cotlnsecx`

`1/v` `(v') = -csc^2xlnsecx + 1`

`v'= v(1-csc^2xlnsecx) =(1-csc^2xlnsecx)secx^cotx`

`dy/dx = (secxtanxlntanx+sec^2xcscx)tanx^secx + (1-csc^2xlnsecx)secx^cotx`