Question

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

Answer 1

`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`