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

Feb 6, 2018

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

#### Explanation:

The quotient rule states that:

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

In this case,

$f \left(x\right) = \sin x \text{ "" "" "=>" "" } f ' \left(x\right) = \cos x$

$g \left(x\right) = x - {e}^{x} \text{ "" "color(white)"-" => " "" } g ' \left(x\right) = 1 - {e}^{x}$

Therefore, if we differentiate with quotient rule, we get:

$\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{x - {e}^{x}}\right) = \frac{\left(x - {e}^{x}\right) \left(\cos x\right) - \left(\sin x\right) \left(1 - {e}^{x}\right)}{x - {e}^{x}} ^ 2$

$= \frac{x \cos x - {e}^{x} \cos x - \sin x + {e}^{x} \sin x}{x - {e}^{x}} ^ 2$

There are other ways to simplify this, but all forms of it are equally correct.