# How do you find the differential dy of the function y=1/3cos((6pix-1)/2)?

Feb 1, 2017

$= - \pi \sin \left(3 \pi x - \frac{1}{2}\right)$

#### Explanation:

$\mathrm{dy} = d \left(\frac{1}{3} \cos \left(\frac{6 \pi x - 1}{2}\right)\right)$

factoring out the constant...
$= \frac{1}{3} d \left(\cos \left(\frac{6 \pi x - 1}{2}\right)\right)$

by the chain rule...
$= \frac{1}{3} \cdot - \sin \left(\frac{6 \pi x - 1}{2}\right) d \left(\frac{6 \pi x - 1}{2}\right)$

executing the chain rule...
$= - \frac{1}{3} \sin \left(\frac{6 \pi x - 1}{2}\right) \frac{6 \pi}{2}$

re-arranging the algebra...
$= - \pi \sin \left(\frac{6 \pi x - 1}{2}\right)$

$= - \pi \sin \left(3 \pi x - \frac{1}{2}\right)$

and as sine is an odd function....
$= \pi \sin \left(\frac{1}{2} - 3 \pi x\right)$