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

1 Answer
Jun 12, 2018

d/dx(ln abs(secx-tanx)) = -secx ddx(ln|secxtanx|)=secx

Explanation:

Using the chain rule:

d/dx(ln abs(secx-tanx)) = 1/(secx-tanx) d/dx (secx-tanx)ddx(ln|secxtanx|)=1secxtanxddx(secxtanx)

d/dx(ln abs(secx-tanx)) = (secxtanx-sec^2x)/(secx-tanx) ddx(ln|secxtanx|)=secxtanxsec2xsecxtanx

d/dx(ln abs(secx-tanx)) = (secx(tanx-secx))/(secx-tanx) ddx(ln|secxtanx|)=secx(tanxsecx)secxtanx

d/dx(ln abs(secx-tanx)) = -secx ddx(ln|secxtanx|)=secx