# Is f(x)=cosx+tanx-sinx increasing or decreasing at x=pi/6?

Mar 19, 2018

Based on the sign of the derivative, we can see that $f \left(x\right)$ is decreasing at $x = \frac{\pi}{6}$

#### Explanation:

We will check if the slope of the curve at $x = \frac{\pi}{6}$ is positive, negative, or zero. This will tell us if the function is increasing (pos.), decreasing (neg.), or steady (zero).

We will find the slope of the curve by taking the derivative of $f \left(x\right)$. Since all the terms are additive, we can take the derivatives individually:

$\frac{\mathrm{dy}}{\mathrm{dx}} f \left(x\right) = f ' \left(x\right) = \frac{d}{\mathrm{dx}} \cos \left(x\right) + \frac{d}{\mathrm{dx}} \tan x - \frac{d}{\mathrm{dx}} \sin x$

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

$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$

$\frac{d}{\mathrm{dx}} \sin x = \cos x$

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

now, we plug in our x-value, $\frac{\pi}{6}$, to find the slope:

$f ' \left(\frac{\pi}{6}\right) = {\sec}^{2} \left(\frac{\pi}{6}\right) - \left(\sin \left(\frac{\pi}{6}\right) + \cos \left(\frac{\pi}{6}\right)\right)$

$f ' \left(\frac{\pi}{6}\right) = 1. \overline{333} - \left(0.5 + 0.8660254\right)$

$f ' \left(\frac{\pi}{6}\right) = 1. \overline{333} - \left(1.366025404\right)$

$\textcolor{red}{f ' \left(\frac{\pi}{6}\right) = - 0.03269}$

We can see that the slope is negative, meaning that the function is $\textcolor{red}{\text{decreasing}}$