What is the derivative of #(tan2x+sec2x)^2#?

Question

1578 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-11T13:07:05

#(dy)/(dx)=4sec2x(tan2x+sec2x)^2#

we will use teh chain rule

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

#y=(tan2x+sec2x)^2#

#u=tan2x+sec2x=>(du)/(dx)=2sec^2 2x+2sec2xtan2x#

#y=u^2=>(dy)/(du)=2u#

#:.(dy)/(dx)=2uxx(2sec^2 2x+2sec2xtan2x)#

substitute back and tidy up

#(dy)/(dx)=2(tan2x+sec2x)(2sec^2 2x+2sec2xtan2x)#

#(dy)/(dx)=4sec2x(tan2x+sec2x)(sec2x+tan2x)#

#(dy)/(dx)=4sec2x(tan2x+sec2x)^2#