# Is f(x)=e^-xcos(-x)-sinx/(pi-e^x) increasing or decreasing at x=pi/6?

$f \left(x\right)$ is decreasing at $x = \setminus \frac{\pi}{6}$

#### Explanation:

Given function:

$f \left(x\right) = {e}^{- x} \setminus \cos \left(- x\right) - \setminus \frac{\setminus \sin x}{\setminus \pi - {e}^{x}}$

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

Differentiating above function w.r.t. $x$ using chain rule & quotient rule as follows

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({e}^{- x} \setminus \cos x + \setminus \frac{\setminus \sin x}{{e}^{x} - \setminus \pi}\right)$

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

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

Setting $x = \setminus \frac{\pi}{6}$ in above expression we get

$f ' \left(\setminus \frac{\pi}{6}\right)$

$= - {e}^{- \setminus \frac{\pi}{6}} \left(\setminus \sin \left(\setminus \frac{\pi}{6}\right) + \setminus \cos \left(\setminus \frac{\pi}{6}\right)\right) + \setminus \frac{{e}^{\setminus \frac{\pi}{6}} \left(\setminus \cos \left(\setminus \frac{\pi}{6}\right) - \setminus \sin \left(\setminus \frac{\pi}{6}\right)\right) - \setminus \pi \setminus \cos \left(\setminus \frac{\pi}{6}\right)}{{\left({e}^{\setminus \frac{\pi}{6}} - \setminus \pi\right)}^{2}}$

$= - 0.80921 - 0.9953$

$= - 1.8045$

Since $f ' \left(\setminus \frac{\pi}{6}\right) < 0$ hence the given function $f \left(x\right)$ is decreasing at $x = \setminus \frac{\pi}{6}$