How do you differentiate # f(x) =(sin x + tan x)/(sin x-cos x) #?

1 Answer
Dec 21, 2015

#f'(x)=((cosx-sec^2x)(sinx-cosx)-(cosx+sinx)(sinx+tanx))/(sinx-cosx)^2#

Explanation:

Original equation:
#f(x)=(sinx+tanx)/(sinx-cosx)#

Use the quotient rule to derive.

Derive the top and times by the bottom:
#(cosx-sec^2x)(sinx-cosx)#

Derive the bottom and multiply by the top:
#(cosx+sinx)(sinx+tanx)#

Subtract the two:
#(cosx-sec^2x)(sinx-cosx)-(cosx+sinx)(sinx+tanx)#

Place it over the bottom squared:
#((cosx-sec^2x)(sinx-cosx)-(cosx+sinx)(sinx+tanx))/(sinx-cosx)^2#

Your final answer should be:
#f'(x)=((cosx-sec^2x)(sinx-cosx)-(cosx+sinx)(sinx+tanx))/(sinx-cosx)^2#