Question

if y = cos t sin 2t then,
what is dy/dt ?

explain with proper steps. ​

Answer 1

Question :-

$\sf{if \: y = \cos t \: \sin2t \: \: then \: find \: \frac{dy}{dt} }$

Answer :-

$\to \boxed{ \sf{ \frac{dy}{dt} = 2 \cos t( \cos2t - { \sin}^{2} t)}} \:$

Explanation :-

Here we will use product rule of differentiation .

That is -

$\to \sf{ \frac{d}{d} \: u.v = u \frac{dv}{dx} + v \frac{du}{dx} }$

here u and v should be function of x .

,we have

$\to \: \sf{y = \cos t \: \sin2t}$

differentiate with respect to "t" on both sides

$\to \sf{ \frac{dy}{dt} = \cos t \frac{d}{dt} \sin2t \: + \sin2t \frac{d}{dt} \cos t} \\ \: \\ \because \boxed{ \sf{ \frac{d}{dx} \sin x = \cos x \:}} \\ \because \boxed{ \sf{ \frac{d}{dx} \cos x = - \sin x}} \\ \therefore \: \\ \to \sf{ \frac{dy}{dt} = \cos t \cos 2t \: \frac{d}{dt}2t + \sin2t \: ( - \sin t) } \\ \\ \to \sf{ \frac{dy}{dt} = 2 \cos2t \cos t - \sin t \: \sin2t \: } \\ \\ \because \boxed{ \sin2 \theta = 2 \sin \theta \cos \theta \: }$

$\therefore \\ \to \sf{ \frac{dy}{dt} = 2 \cos2t \: \cos t \: - 2 \cos t \: { \sin}^{2} t \: } \\ \\ \to \boxed{ \sf{ \frac{dy}{dt} = 2 \cos t( \cos2t - { \sin}^{2} t)}}$

Answer 2

Answer:

yo chets how r u? long time no see