Question

differentiation of cos omega t?

Answer 1

Differentiation of $\cos(\omega t)$:

• As per question , the equation is :

$y = \cos( \omega t)$

• Now, differentiation with respect to t :

$\implies \: \dfrac{dy}{dt} = \dfrac{d \{ \cos( \omega t) \} }{dt}$

• Applying chain rule :

$\implies \: \dfrac{dy}{dt} = \dfrac{d \{ \cos( \omega t) \} }{d( \omega t)} \times \dfrac{d( \omega t)}{dt}$

$\implies \: \dfrac{dy}{dt} = - \sin( \omega t) \times \omega$

$\boxed{ \implies \: \dfrac{dy}{dt} = - \omega \sin( \omega t) }$

Hope It Helps.