How do you differentiate $f (x) = \frac{sin x + tan x}{sin x - cos x}$ $f(x) =(sin x + tan x)/(sin x-cos x)$ ?

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Answer 1 · 2015-12-21T04:10:42

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

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

How do you differentiate f(x) =(sin x + tan x)/(sin x-cos x) f(x)=sinx+tanxsinx−cosx f(x) =(sin x + tan x)/(sin x-cos x) ?