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

Dec 2, 2015

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

#### Explanation:

A function is increasing or decreasing at a given point if its first derivative, evaluated at that point, is positive or negative respectively.

To answer, then, we first must find the derivative of $f \left(x\right)$, then evaluate it at $x = \frac{\pi}{6}$, and finally check the sign of the result.

First, let's use the quotient rule to evaluate the derivative:
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \cos \frac{x}{e} ^ x$

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

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

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

Next, we evaluate the first derivative at $x = \frac{\pi}{6}$

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

$= - \frac{\frac{\sqrt{3}}{2} + \frac{1}{2}}{e} ^ \left(\frac{\pi}{6}\right)$

$= \frac{- \sqrt{3} - 1}{2 {e}^{\frac{\pi}{6}}}$

Finally, we check the sign of the result.

$- \sqrt{3} - 1 < 0$ and $2 {e}^{\frac{\pi}{6}} > 0$

thus, as the quotient of a positive number and a negative number is negative,

$f ' \left(\frac{\pi}{6}\right) < 0$

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