If sin(x+y) = log(x+y),then frac{dy}{dx} = | vedclass

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(D) Given the implicit function $\sin(x+y) = \log(x+y)$.Let $u = x+y$. Then the equation becomes $\sin(u) = \log(u)$.Differentiating both sides with r

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$-1$

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(D) Given the implicit function $\sin(x+y) = \log(x+y)$.Let $u = x+y$. Then the equation becomes $\sin(u) = \log(u)$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(\sin(u)) = \frac{d}{dx}(\log(u))$$\cos(u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$Since $\frac{du}{dx} = \frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}$,we have:$\cos(u) \cdot (1 + \frac{dy}{dx}) = \frac{1}{u} \cdot (1 + \frac{dy}{dx})$$(\cos(u) - \frac{1}{u}) \cdot (1 + \frac{dy}{dx}) = 0$Since $\sin(u) = \log(u)$,$\cos(u) 
eq \frac{1}{u}$ in general,therefore:$1 + \frac{dy}{dx} = 0$$\frac{dy}{dx} = -1$.

If $\sin(x+y) = \log(x+y)$,then $\frac{dy}{dx} =$

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