# How do you find the derivative of  (pi/2)sinx-cosx?

Jun 24, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\pi}{2}\right) \cos x + \sin x$

#### Explanation:

$y = \left(\frac{\pi}{2}\right) \sin x - \cos x$

Since $\frac{\pi}{2}$ is a constant we can apply linearity for the derivative.

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\pi}{2}\right) \frac{d}{\mathrm{dx}} \left(\sin x\right) - \frac{d}{\mathrm{dx}} \left(\cos x\right)$

Applying standard derivatives

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\pi}{2}\right) \cos x - \left(- \sin x\right)$

$= \left(\frac{\pi}{2}\right) \cos x + \sin x$