If (a+sqrt{2} b cos x)(a-sqrt{2} b cos y)=a^2 | vedclass

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Question

(B) Given the equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$. Differentiating both sides with respect to $x$: $(a+\sqrt{2} b \cos x) \

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$\frac{a-b}{a+b}$

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(B) Given the equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$. Differentiating both sides with respect to $x$: $(a+\sqrt{2} b \cos x) \cdot (\sqrt{2} b \sin y \frac{dy}{dx}) + (a-\sqrt{2} b \cos y) \cdot (-\sqrt{2} b \sin x) = 0$. At the point $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,we have $\cos x = \cos y = \frac{1}{\sqrt{2}}$ and $\sin x = \sin y = \frac{1}{\sqrt{2}}$. Substituting these values: $(a+\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) \cdot (\sqrt{2} b \cdot \frac{1}{\sqrt{2}} \cdot \frac{dy}{dx}) + (a-\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) \cdot (-\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) = 0$. $(a+b) \cdot (b \frac{dy}{dx}) + (a-b) \cdot (-b) = 0$. $(a+b) b \frac{dy}{dx} = b(a-b)$. Since $b > 0$,we can divide by $b$: $(a+b) \frac{dy}{dx} = a-b$. Therefore,$\frac{dy}{dx} = \frac{a-b}{a+b}$.

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$ where $a>b>0$,then at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,$\frac{dy}{dx}=$

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