How do I differentiate #y= sec^2(x) + tan^2(x)#?

Question

Luke Alvarez · Accepted Answer

Use the chain rule (generalized power rule) and the derivatives of the trigonometric functions.  Beyond that, there are choices you can make Choice 1:
It may help to re-write it as:
#y=(secx)^2+(tanx)^2#.  So #(dy)/(dx)=2(secx)*d/(dx)(secx) + 2(tanx)*d/(dx)(tan x)# Thus< #(dy)/(dx)=2secx*secxtanx+2tanx*sec^2x# #(dy)/(dx)=4sec^2xtanx#. Which, since #sec^2x=tan^2x+1#, could also be written  #(dy)/(dx)=4tan^3x+4tanx# Choice 2:
Use the trigonometric identity to re-write the function using #tanx#:
#y=sec^2x+tan^2x=(tan^2x+1)+tan^2x=2tan^2x+1# So,
#(dy)/(dx)=4tanxd/(dx)(tanx)=4tanxsec^2x# Choice 3:
Use the trigonometric identity just mentioned to re-write the function using #secx#: #y=sec^2x+tan^2x=sec^2x+sec^2x-1=2sec^x-1#
So,
#(dy)/(dx)=4secxd/(dx)(secx)=4secxsecxtanx=4sec^2xtanx#

Scarlett Allison · Answer

undefined To differentiate $ y = \sec^2(x) + \tan^2(x) $, use the chain rule for differentiation:

$$ \frac{d}{dx} \left( \sec^2(x) + \tan^2(x) \right) = \frac{d}{dx} \sec^2(x) + \frac{d}{dx} \tan^2(x) $$

Apply the derivative formulas for $ \sec^2(x) $ and $ \tan^2(x) $:

$$ \frac{d}{dx} \sec^2(x) = 2\sec(x) \tan(x) \sec(x) = 2\sec(x) \tan(x) $$

$$ \frac{d}{dx} \tan^2(x) = 2\tan(x) \sec^2(x) $$

Substitute these derivatives back into the original expression:

$$ \frac{d}{dx} \left( \sec^2(x) + \tan^2(x) \right) = 2\sec(x) \tan(x) + 2\tan(x) \sec^2(x) $$

$$ = 2\sec(x) \tan(x) + 2\tan(x) \sec^2(x) $$