Question

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

Answer 1

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

Explanation:

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`