How do you find the derivative of $cosx/(sinx-2)$ ?

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Answer 1 · 2015-03-30T18:29:02

The answer is: $y'=(2sinx-1)/(sinx-2)^2$

This is because:

$y'=(-sinx*(sinx-2)-cosx*cosx)/(sinx-2)^2=$

$=(-sin^2x+2sinx-cos^2x)/(sinx-2)^2=(2sinx-1)/(sinx-2)^2$

This is because: $sin^2x+cos^2x=1$ .

How do you find the derivative of cosx/(sinx-2)cosx/(sinx-2)?