If sin y + e^{-x cos y} = e, then frac{dy}{dx | vedclass

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(C) Given the equation: $\sin y + e^{-x \cos y} = e$. Differentiating both sides with respect to $x$ using the chain rule: $\cos y \frac{dy}{dx} + e^{

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$e$

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(C) Given the equation: $\sin y + e^{-x \cos y} = e$. Differentiating both sides with respect to $x$ using the chain rule: $\cos y \frac{dy}{dx} + e^{-x \cos y} \cdot \frac{d}{dx}(-x \cos y) = 0$ $\cos y \frac{dy}{dx} + e^{-x \cos y} \cdot [(-1) \cos y + (-x)(-\sin y) \frac{dy}{dx}] = 0$ $\cos y \frac{dy}{dx} + e^{-x \cos y} [-\cos y + x \sin y \frac{dy}{dx}] = 0$ $\frac{dy}{dx} (\cos y + x \sin y e^{-x \cos y}) = \cos y e^{-x \cos y}$ $\frac{dy}{dx} = \frac{\cos y e^{-x \cos y}}{\cos y + x \sin y e^{-x \cos y}}$ Now,substitute $(x, y) = (1, \pi)$: $\cos \pi = -1$ and $\sin \pi = 0$. $\frac{dy}{dx} \Big|_{(1, \pi)} = \frac{(-1) e^{-1(-1)}}{-1 + 1(0) e^{-1(-1)}} = \frac{-e}{-1} = e$.

If $\sin y + e^{-x \cos y} = e,$ then $\frac{dy}{dx}$ at $(1, \pi)$ is

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