Question

How do you find the derivative of `y = [(tanx - 1) / secx]`?

Answer 1

`dy/dx=cosx+sinx`

Explanation:

We need to know the following derivatives:

• `d/dx(sinx)=cosx`
• `d/dx(cosx)=-sinx`

First we can simplify the function using `tanx=sinx/cosx` and `secx=1/cosx`:

`y=(sinx/cosx-1)/(1/cosx)=cosx(sinx/cosx-1)=sinx-cosx`

Then the derivative is:

`dy/dx=d/dx(sinx)-d/dx(cosx)`

`dy/dx=cosx-(-sinx)`

`dy/dx=cosx+sinx`