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

1 Answer
Jan 29, 2017

#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#