Title: Find the derivative of x⋅siny+y⋅cosx=0 Post by: bio_man on Sep 30, 2018 \(x\cdot \sin y+y\cdot \cos x=0\)
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: \(\frac{dy}{dx}=\frac{y\cdot \sin x-\sin y}{x\cdot \cos y+\cos x}\) Title: Re: Find the derivative of Post by: duddy on Nov 7, 2019 👍
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