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

Feb 7, 2016

The function is decreasing.

#### Explanation:

To figure the increasing or decreasing nature of a function we can look at its derivative. If $f ' \left(x\right)$ is positive at the specified $x$ value then the function is increasing otherwise if $f ' \left(x\right)$ is negative at the specified value then the function is decreasing.

If the $f ' \left(x\right) = 0$ then we have a stationary point.

So, to differentiate the function:

$f ' \left(x\right) = - {\csc}^{2} \left(x\right) - {e}^{x} \tan x - {e}^{x} {\sec}^{2} \left(x\right)$

The product rule was used in the differentiation.

Now putting our value of $x$ into our function:

$f ' \left(\frac{\pi}{6}\right) = - {\csc}^{2} \left(\frac{\pi}{6}\right) - {e}^{\frac{\pi}{6}} \tan \left(\frac{\pi}{6}\right) - {e}^{\frac{\pi}{6}} {\sec}^{2} \left(\frac{\pi}{6}\right)$

$= - 4 - {e}^{\frac{\pi}{6}} \left(\frac{1}{3} + \frac{4}{3}\right) = - 4 - \frac{5}{4} {e}^{\frac{\pi}{6}} < 0$

Therefore as $f ' \left(\frac{\pi}{6}\right)$ is negative our function must be decreasing as we can see from the graph of $f \left(x\right)$ below.