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

Question

5016 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-05T08:31:32

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

Here ,

#y=ln(sec^2x)#

Let ,

#y=lnu , where , u=sec^2x#

#(dy)/(du)=1/u and (du)/(dx)=2secx*secxtanx=2sec^2xtanx#

Diff.w.r.t. #x# using Chain Rule:

#color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)#

#(dy)/(dx)=1/u xx 2sec^2xtanx#

Subst, back , #u=sec^2x#

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

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