Find frac{dy}{dx} for the equation 2x + 3y = | vedclass

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Question

(A) The given equation is $2x + 3y = \sin y$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{d

Vedclass · Accepted Answer

$\frac{2}{\cos y - 3}$

Vedclass · Answer

(A) The given equation is $2x + 3y = \sin y$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$ $2 + 3 \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$ Rearranging the terms to isolate $\frac{dy}{dx}$: $2 = \cos y \cdot \frac{dy}{dx} - 3 \frac{dy}{dx}$ $2 = (\cos y - 3) \frac{dy}{dx}$ Therefore,$\frac{dy}{dx} = \frac{2}{\cos y - 3}$.

Find $\frac{dy}{dx}$ for the equation $2x + 3y = \sin y$.

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