Question

What is the derivative of `f(x)=sec^2xcosx`?

Answer 1

`dy/dx = tan(x)sec(x)`

Explanation:

`y = sec^2(x)cos(x)`

But since `sec(x) = 1/cos(x)` we can simplify that to

`y = sec(x)` or `y = 1/cos(x)`

from there, using either the product + chain rule or the quotient rule

`dy/dx = d/(du)1/u*d/dxcos(x)`
`dy/dx = -1/u^2 *(-sin(x))`
`dy/dx = sin(x)/cos^2(x)`
`dy/dx = sin(x)/cos(x)*1/cos(x) = tan(x)sec(x)`