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3 months ago


\(x\cdot \sin y+y\cdot \cos x=0\)

Rewrite with brackets for organization purposes:

\(\left(x\cdot \sin y\right)+\left(y\cdot \cos x\right)=0\)

Product rule twice, also differentiate implicitly with respect to \(x\):

\(\left(\sin y+x\cos y\ \frac{dy}{dx}\right)+\left(\frac{dy}{dx}\cos x-y\cdot \sin x\right)=0\)

Isolate for dy/dx:

\(\left(x\cos y\ \frac{dy}{dx}\right)+\left(\frac{dy}{dx}\cos x\right)=-\sin y+y\cdot \sin x\)

Factor out dy/dx on the left side:

\(\frac{dy}{dx}\left(x\cdot \cos y+\cos x\right)=-\sin y+y\cdot \sin x\)

Divide both sides by: \(\left(x\cdot \cos y+\cos x\right)\), giving us:

\(\frac{dy}{dx}=\frac{y\cdot \sin x-\sin y}{x\cdot \cos y+\cos x}\)
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