Question

What is the derivative of `y= ln abs(secx-tanx)` for `x>0`?

Answer 1

`d/dx(ln abs(secx-tanx)) = -secx`

Explanation:

Using the chain rule:

`d/dx(ln abs(secx-tanx)) = 1/(secx-tanx) d/dx (secx-tanx)`

`d/dx(ln abs(secx-tanx)) = (secxtanx-sec^2x)/(secx-tanx)`

`d/dx(ln abs(secx-tanx)) = (secx(tanx-secx))/(secx-tanx)`

`d/dx(ln abs(secx-tanx)) = -secx`