Question

How do you take the derivative of `y=sec^2 x + tan^2 x`?

Answer 1

`y^' = 4sec^2(x)tan(x)`

Explanation:

Assuming that you're aware of the fact that

`color(blue)(d/dx(tanx) = sec^2x)" "` and `" "color(blue)(d/dx(secx) = secx * tanx)`

you can differentiate this function by using the chain rule twice, once for `u^2`, with `u = secx` and once for `t^2`, with `t = tanx`.

This will get you

`d/dx(y) = d/dx(sec^2x) + d/dx(tan^2x)`

`y^' = d/(du)(u^2) * d/dx(u) + d/(dt)t^2 * d/dx(t)`

`y^' = 2u * d/dx(secx) + 2t * d/dx(tanx)`

`y^' = 2sec(x) * sec(x) * tan(x) + 2tan(x) * sec^2(x)`

This is equal to

`y^' = 2sec^2(x)tan(x) + 2sec^2(x)tan(x)`

`y^' = color(green)(4sec^2(x)tan(x))`