How do you differentiate ln(sec^2 * x)ln(sec2x)?

1 Answer
Jul 5, 2018

(dy)/(dx)=2tanxdydx=2tanx

Explanation:

Here ,

y=ln(sec^2x)y=ln(sec2x)

Let ,

y=lnu , where , u=sec^2xy=lnu,where,u=sec2x

(dy)/(du)=1/u and (du)/(dx)=2secx*secxtanx=2sec^2xtanxdydu=1uanddudx=2secxsecxtanx=2sec2xtanx

Diff.w.r.t. xx using Chain Rule:

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)dydx=dydududx

(dy)/(dx)=1/u xx 2sec^2xtanxdydx=1u×2sec2xtanx

Subst, back , u=sec^2xu=sec2x

:.(dy)/(dx)=1/sec^2x xx 2sec^2xtanx

(dy)/(dx)=2tanx