How do you differentiate f(x) = (sinx)/(sinx-cosx) using the quotient rule?

Jan 13, 2016

$f ' \left(x\right) = - \cos x \frac{\sin x + \cos x}{1 - \sin 2 x}$

Explanation:

The quotent rule states that:

$a \left(x\right) = \frac{b \left(x\right)}{c \left(x\right)}$

Then:

$a ' \left(x\right) = \frac{b ' \left(x\right) \cdot c \left(x\right) - b \left(x\right) \cdot c ' \left(x\right)}{c \left(x\right)} ^ 2$

Likewise for $f \left(x\right)$:

$f \left(x\right) = \frac{\sin x}{\sin x - \cos x}$

$f ' \left(x\right) = \frac{\left(\sin x\right) ' \left(\sin x - \cos x\right) - \sin x \left(\sin x - \cos x\right) '}{\sin x - \cos x} ^ 2$

$f ' \left(x\right) = \frac{\cos x \left(\sin x - \cos x\right) - \sin x \left(\cos x - \left(- \cos x\right)\right)}{\sin x - \cos x} ^ 2$

$f ' \left(x\right) = \frac{\cos x \sin x - {\cos}^{2} x - \sin x \cos x - \sin x \cos x}{\sin x - \cos x} ^ 2$

$f ' \left(x\right) = \frac{- \sin x \cos x - {\cos}^{2} x}{\sin x - \cos x} ^ 2$

$f ' \left(x\right) = - \cos x \frac{\sin x + \cos x}{\sin x - \cos x} ^ 2$

$f ' \left(x\right) = - \cos x \frac{\sin x + \cos x}{{\sin}^{2} x - 2 \sin x \cos x + {\cos}^{2} x}$

$f ' \left(x\right) = - \cos x \frac{\sin x + \cos x}{\left({\sin}^{2} x + {\cos}^{2} x\right) - 2 \sin x \cos x}$

$f ' \left(x\right) = - \cos x \frac{\sin x + \cos x}{1 - \sin 2 x}$