Consider the equation $x \sin y + 2 y = 6$ .Assuming this implicitly defines y as a function of x, find $\frac { d y } { d x }$ .
A) $- \frac { \sin y } { x \cos y + 2 }$
B) $- \frac { 6 \sin y } { x \cos y + 2 }$
C) $- \frac { \sin y + 2 } { x \cos y }$
D) $- \frac { \sin y + 6 } { x \cos y }$

Question

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we differentiate both sides of the equation $x \sin y + 2 y = 6$ with respect to $x$, treating $y$ as a function of $x$. Using the product rule and chain rule, we get:$$x \cos y \frac{dy}{dx} + \sin y + 2\frac{dy}{dx} = 0$$Rearranging to solve for $\frac{dy}{dx}$, we get:$$\frac{dy}{dx} = -\frac{\sin y}{x \cos y + 2}$$This matches choice A.

Consider the Equation xsin⁡y+2y=6x \sin y + 2 y = 6xsiny+2y=6

Consider the Equation $x \sin y + 2 y = 6$