What is the derivative of #y=sec^2 x + tan^2 x#?

2 Answers
Jun 23, 2018

#f'(x)=2sec^2(x)tan(x)+2tan(x)*(1+tan^2(x))#

Explanation:

Note that

#(sec(x))'=sec(x)tan(x)#
#(tan(x))'=1+tan^2(x)#

Jun 23, 2018

Shown below...

Explanation:

Another way of thinking about it:

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

#=>1 + tan^2x + tan^2 x = sec^2 x + tan^2 x #

#=> 1 + 2tan^2 x = sec^2 x + tan^2 x #

#=> d/dx ( 1 + 2 tan^2 x ) #

Use #d/dx tanx = sec^2 x #

Use chain rule:

#= 2*2tanx * d/dx tanx #

# = 4tanx sec^2x #