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

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(A) Given the equation $3\sin(xy) + 4\cos(xy) = 5$. Let $u = xy$. Then $3\sin(u) + 4\cos(u) = 5$. Differentiating both sides with respect to $x$: $3\c

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$-\frac{y}{x}$

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(A) Given the equation $3\sin(xy) + 4\cos(xy) = 5$. Let $u = xy$. Then $3\sin(u) + 4\cos(u) = 5$. Differentiating both sides with respect to $x$: $3\cos(u) \cdot \frac{du}{dx} - 4\sin(u) \cdot \frac{du}{dx} = 0$. $(3\cos(u) - 4\sin(u)) \cdot \frac{du}{dx} = 0$. Since $3\sin(u) + 4\cos(u) = 5$,the expression $3\cos(u) - 4\sin(u)$ cannot be zero for all $x$. Thus,$\frac{du}{dx} = 0$. Since $u = xy$,we have $\frac{d}{dx}(xy) = 0$. Using the product rule: $y + x \cdot \frac{dy}{dx} = 0$. Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

If $3\sin (xy) + 4\cos (xy) = 5$,then $\frac{dy}{dx} = $

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