# How do you differentiate f(x) = (cos^2x)/(e^(x-2)+x) using the quotient rule?

The quotient rule is that, if you have $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then $f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h {\left(x\right)}^{2}}$.
Here, we will take $g \left(x\right) = {\cos}^{2} \left(x\right)$, and $h \left(x\right) = {e}^{x - 2} + x$
$g ' \left(x\right) = - 2 \sin x \cos x$
$h ' \left(x\right) = {e}^{x - 2} + 1$
So, $f ' \left(x\right) = \frac{\left({e}^{x - 2} + x\right) \left(- 2 \sin x \cos x\right) - \left({e}^{x + 2} + 1\right) \left({\cos}^{2} x\right)}{{e}^{x + 2} + x} ^ 2 = - \frac{\cos x \left({e}^{x - 2} + x\right) \left(2 \sin x\right) - \left({e}^{x + 2} + 1\right) \left(\cos x\right)}{{e}^{x + 2} + x} ^ 2$