How do you differentiate #y=sec^2x+tan^2x#?

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Answer 1 · 2018-07-27T06:51:38

#(dy)/(dx)=4sec^2xtanx#

We know that,

#color(red)((1)d/(dx)(secx)=secxtanx#

#color(green)((2)d/(dx)(tanx)=sec^2x#

Here,

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

Diff.w.r.t. #x# ,#"using "color(blue)"Chain Rule :"#

#(dy)/(dx)=2secxcolor(red)(d/(dx)(secx))+2tanx color(green)(d/(dx)(tanx)#

#:.(dy)/(dx)=2secx*color(red)(secxtanx)+2tanx*color(green)(sec^2x#

#:.(dy)/(dx)=2sec^2xtanx+2sec^2xtanx#

#:.(dy)/(dx)=4sec^2xtanx#