# Is f(x)=5 cos(3x+(5pi)/6)+ 2sin(4x-(3pi)/4)  increasing or decreasing at x=pi/12 ?

May 22, 2018

Apply the Chain Rule to find the first derivative, then plug in $\frac{\pi}{12}$.

#### Explanation:

$f \left(x\right) = 5 \cos \left(3 x + 5 \frac{\pi}{6}\right) + 2 \sin \left(4 x - 3 \frac{\pi}{4}\right)$

$f ' \left(x\right) = - 5 \sin \left(3 x + 5 \frac{\pi}{6}\right) \cdot 3 + 2 \cos \left(4 x - 3 \frac{\pi}{4}\right) \cdot 4$

$f ' \left(x\right) = - 15 \sin \left(3 x + 5 \frac{\pi}{6}\right) + 8 \cos \left(4 x - 3 \frac{\pi}{4}\right)$

Plug in $\frac{\pi}{12.}$

$f ' \left(\frac{\pi}{12}\right) = - 15 \sin \left(\frac{\pi}{4} + 5 \frac{\pi}{6}\right) + 8 \cos \left(\frac{\pi}{3} - 3 \frac{\pi}{4}\right)$

$f ' \left(\frac{\pi}{12}\right) = - 15 \sin \left(13 \frac{\pi}{12}\right) + 8 \cos \left(- 5 \frac{\pi}{12}\right)$

I don't know the unit circle down to the twelfths, so I just used my calculator to do this whole thing.

$f ' \left(\frac{\pi}{12}\right) = 3.8823 + 2.0706$

$f ' \left(\frac{\pi}{12}\right) = 5.9529$

If the problem strictly asked whether it was increasing or decreasing, it's increasing, because the value of the first derivative at that point is positive. The slope is 5.9529 or whatever we calculated.