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Educator
Posts: 24713
9 months ago
 Find the derivative of $$x\cdot \sin y+y\cdot \cos x=0$$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:$$\frac{dy}{dx}=\frac{y\cdot \sin x-\sin y}{x\cdot \cos y+\cos x}$$ Read 240 times
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