$$\text { Find } \frac { d y } { d x }$$
-$\ln y = e ^ { y } \cos 3 x$
A) $- 3 y e ^ { y } \sin 3 x$
B) $e ^ { y } \cos 3 x - 3 e ^ { y } \sin 3 x$
C) $\frac { y e ^ { y } \sin 3 x } { 1 - e ^ { y } \cos 3 x }$
D) $\frac { - 3 y e ^ { y } \sin 3 x } { 1 - y e ^ { y } \cos 3 x }$

Question

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we differentiate both sides of the equation $\ln y = e^y \cos 3x$ with respect to $x$, using implicit differentiation. On the left side, using the chain rule, we get $\frac{1}{y} \frac{dy}{dx}$. On the right side, applying the product rule and chain rule, we get $e^y \cos 3x \frac{dy}{dx} - 3e^y \sin 3x$. Setting these equal gives $\frac{1}{y} \frac{dy}{dx} = e^y \cos 3x \frac{dy}{dx} - 3e^y \sin 3x$. Solving for $\frac{dy}{dx}$ yields $\frac{dy}{dx} = \frac{-3ye^y \sin 3x}{1 - ye^y \cos 3x}$.

Find dydx\text { Find } \frac { d y } { d x } Find dxdy​

$\text { Find } \frac { d y } { d x }$