$\text { Differentiate } 3 \sin x \cos y = 1 \text { with respect to } t ( x \text { and } y \text { are functions of } t ) \text {. }$
A) $\sin x \cos y \frac { d x } { d t } = \cos x \sin y \frac { d y } { d t }$
B) $\cos x \cos y \frac { d x } { d t } = \sin x \sin y \frac { d y } { d t }$
C) $x \sin y \frac { d x } { d t } = \sin x \cos y \frac { d y } { d t }$
D) $\cos x \cos y \frac { d x } { d t } = 3 \sin x \sin y \frac { d y } { d t }$
E) $3 \sin x \sin y \frac { d x } { d t } = \cos x \cos y \frac { d y } { d t }$

Question

$\text { Differentiate } 3 \sin x \cos y = 1 \text { with respect to } t ( x \text { and } y \text { are functions of } t ) \text {. }$ 
A) $\sin x \cos y \frac { d x } { d t } = \cos x \sin y \frac { d y } { d t }$
B) $\cos x \cos y \frac { d x } { d t } = \sin x \sin y \frac { d y } { d t }$
C) $x \sin y \frac { d x } { d t } = \sin x \cos y \frac { d y } { d t }$
D) $\cos x \cos y \frac { d x } { d t } = 3 \sin x \sin y \frac { d y } { d t }$
E) $3 \sin x \sin y \frac { d x } { d t } = \cos x \cos y \frac { d y } { d t }$

Quizplus · Accepted Answer

Taking the derivative with respect to t on both sides of the equation:
$$\frac{d}{dt} (3 \sin x \cos y) = \frac{d}{dt} (1)$$
Using the product rule of differentiation, we get:
$$3\cos y \frac{dx}{dt} \sin y + 3\sin x (-\sin y) \frac{dy}{dt} = 0$$
Dividing both sides by 3 and rearranging, we get:
$$\cos x \cos y \frac{dx}{dt} = \sin x \sin y \frac{dy}{dt}$$
Therefore, the correct choice is B.

Differentiate 3sin⁡xcos⁡y=1 with respect to t(x and y are functions of t). \text { Differentiate } 3 \sin x \cos y = 1 \text { with respect to } t ( x \text { and } y \text { are functions of } t ) \text {. } Differentiate 3sinxcosy=1 with respect to t(x and y are functions of t).

$\text { Differentiate } 3 \sin x \cos y = 1 \text { with respect to } t ( x \text { and } y \text { are functions of } t ) \text {. }$