The tutor shows how to remind yourself that the derivative of tanx is sec²x.

Let’s imagine you don’t recall that (tanx)’=sec²x. Here’s how to reconstruct it:

1. Recall that tanx=sinx/cosx
2. Take the derivative of sinx/cosx using the quotient rule:(u/v)’ = (vu’-uv’)/v²
3. In this case, (sinx/cosx)’ = [cosx(cosx) – sinx(-sinx)]/cos²x
4. Recall that cos²x +sin²x = 1.
5. Simplifying we arrive at (sinx/cosx)’ = 1/cos²x = sec²x

Therefore, (tanx)’=(sinx/cosx)’=sec²x.

Source:

Larson, Roland and Robert Hostetler. Calculus, part one. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.