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

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Answer 1 · 2017-01-29T23:17:28

$dy/dx=cosx+sinx$

We need to know the following derivatives:

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:

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

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

$dy/dx=cosx+sinx$

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