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

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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 $y$ using the product

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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 $y$ using the product rule: $(a + \sqrt{2} b \cos x) \cdot \frac{d}{dy}(a - \sqrt{2} b \cos y) + (a - \sqrt{2} b \cos y) \cdot \frac{d}{dy}(a + \sqrt{2} b \cos x) = 0$ $(a + \sqrt{2} b \cos x)(\sqrt{2} b \sin y) + (a - \sqrt{2} b \cos y)(-\sqrt{2} b \sin x \frac{dx}{dy}) = 0$ Rearranging to solve for $\frac{dx}{dy}$: $\frac{dx}{dy} = \frac{\sqrt{2} b \sin y (a + \sqrt{2} b \cos x)}{\sqrt{2} b \sin x (a - \sqrt{2} b \cos y)}$ At the point $(\frac{\pi}{4}, \frac{\pi}{4})$,we have $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$: $\frac{dx}{dy} = \frac{\sqrt{2} b (\frac{1}{\sqrt{2}}) (a + \sqrt{2} b \cdot \frac{1}{\sqrt{2}})}{\sqrt{2} b (\frac{1}{\sqrt{2}}) (a - \sqrt{2} b \cdot \frac{1}{\sqrt{2}})}$ $\frac{dx}{dy} = \frac{b(a + b)}{b(a - b)} = \frac{a + b}{a - b}$

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

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