Question

Differentiate y w.r.t x y=AsinWx + AcosWx

Answer 1

$y = A \sin( \omega x) + A \cos( \omega x)$

$= > \dfrac{dy}{dx} = \dfrac{d \{A \sin( \omega x) + A \cos( \omega x) \}}{dx}$

$= > \dfrac{dy}{dx} = \dfrac{d \{A \sin( \omega x) \}}{dx} + \dfrac{d \{A \cos( \omega t) \} }{dx}$

Applying Chain rule , we get :

$= > \dfrac{dy}{dx} = A \omega \cos( \omega t) - A \omega \sin( \omega t)$

$= > \dfrac{dy}{dx} = A \omega \{\cos( \omega t) - \sin( \omega t) \}$

So final answer :

$\boxed{ \red{ \bold{ \dfrac{dy}{dx} = A \omega \{\cos( \omega t) - \sin( \omega t) \}}}}$

Basic formulas for differentiation used here :

$1) \: \: \: f(x) = \cos(x)$

$= > \dfrac{d \{f(x) \}}{dx} = - \sin(x)$

$2) \: \: \: f(x) = \sin(x)$

$= > \dfrac{d \{f(x) \}}{dx} = \cos(x)$