How to differentiate y= (sec x + tan x)(sec x - tan x) ?

1 Answer
Apr 18, 2018

#(dy)/(dx)=0#

Explanation:

Here,

#y=(secx+tanx)(secx-tanx)#

Using product rule,

#(dy)/(dx)=(secx+tanx)d/(dx)((secx-tanx))+(secx-tanx)d/(dx) ((secx+tanx)#

#=(secx+tanx)(secxtanx-sec^2x)+(secx-tanx)(secxtanx+sec^2x)#

#=color(red)(sec^2xtanx)-cancelsec^3x+cancel(secxtan^2x)- color(red)(sec^2xtanx)+color(blue)(sec^2xtanx)+cancelsec^3x- cancel(secxtan^2x)-color(blue)(sec^2xtanx#

#=>(dy)/(dx)=0#

OR

#y=(secx+tanx)(secx-tanx)#

#y=sec^2x-tan^2x#

#y=1#

#=>(dy)/(dx)=0#