Question

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

Answer 1

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`