$(m-n\cos y)(m+n\cos x)=m^2 -n^2$ then find $\frac{dy}{dx}$
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Suppose that you have a relation between $y$ and $x$, say $F(x,y)=0$ such that you can always find $y$ as a function of $x$. Then it can be proved that: $$\frac{dy}{dx}=\frac{-F_x}{F_y}$$ wherein $F_x$ means the derivative of $F$ regarding to variable $x$. Here put $m^2 -n^2=a$ as a constant and so $a'=0$. We have: $$F(x,y)=(m-n\cos y)(m+n\cos x)-a=0$$ then $$F(x,y)=m^2+mn\cos x-mn\cos y-n^2\cos x\cos y-a=0$$ then $$F_x(x,y)=-mn\sin x+n^2\sin x\cos y$$ and $$F_y(x,y)=mn\sin y+n^2\cos x\sin y$$ |
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$$(m-n\cos y)(m+n\cos x)=m^2 -n^2$$ $$(m-n\cos y)(-n\sin x) + (n\sin y)\bigg(\frac{dy}{dx}\bigg)(m+n\cos x)=0$$ $$\frac{dy}{dx}=\frac{(m-n\cos y)(n\sin x)}{(n\sin y)(m+n\cos x)}$$ $$\frac{dy}{dx}=\frac{(m-n\cos y)(\sin x)}{(\sin y)(m+n\cos x)}$$ |
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@Alex has already showed you the steps. Use implicit differentiation and you'll get dy/dx, an then just solve for dy/dx. |
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From $(m-n\cos y)(m+n \cos x)=m^2-n^2$ you can get $\cos y$. $$ n(m+n\cos x)\cos y= n^2+mn\cos x\\ (m+n\cos x)\cos y = n+m\cos x\\ \cos y = \frac{n+m\cos x}{m+n\cos x}\\ y(x)=\arccos \left(\frac{n+m\cos x}{m+n\cos x}\right) $$ Then rest is an exercise in computing derivative. (Rather tedious, unless you notice some trick. I did not notice an easy way to do this.) Here's what WA says. WolframAlpha gives $\csc(x) \sqrt{\frac{(m^2-n^2)\sin^2(x)}{(m+n\cos x)^2}}$ as a result, which is $\pm \frac{\sqrt{m^2-n^2}}{m+n\cos x}$. It seems close to one of your options. |
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