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 $\sin \pi = 0$. $\frac{dy}{dx} \Big|_{(1, \pi)} = \frac{(-1) e^{-1(-1)}}{-1 + 1(0) e^{-1(-1)}} = \frac{-e}{-1} = e$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

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

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