If 3 sin xy + 4 cos xy = 5,then frac{dy}{dx} | vedclass

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(C) Given the equation $3 \sin xy + 4 \cos xy = 5$. Let $xy = t$. Differentiating $xy = t$ with respect to $x$ using the product rule,we get $x \frac{

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$\frac{-y}{x}$

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(C) Given the equation $3 \sin xy + 4 \cos xy = 5$. Let $xy = t$. Differentiating $xy = t$ with respect to $x$ using the product rule,we get $x \frac{dy}{dx} + y = \frac{dt}{dx} \dots (I)$. Now,differentiate the given equation $3 \sin t + 4 \cos t = 5$ with respect to $t$: $\frac{d}{dt}(3 \sin t + 4 \cos t) = \frac{d}{dt}(5)$ $3 \cos t - 4 \sin t = 0$. Since $3 \cos t - 4 \sin t = 0$,it implies $\frac{dt}{dx} = 0$. Substituting this into equation $(I)$,we get $x \frac{dy}{dx} + y = 0$. Therefore,$\frac{dy}{dx} = \frac{-y}{x}$.

If $3 \sin xy + 4 \cos xy = 5$,then $\frac{dy}{dx}$ is equal to . . . . . .

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