If y=1+xe^y,then frac{dy}{dx}= | vedclass

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Question

(A) Given the equation: $y = 1 + xe^y$ Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} = 0 + \left( x \cdot e^y

Vedclass · Accepted Answer

$\frac{e^y}{2-y}$

Vedclass · Answer

(A) Given the equation: $y = 1 + xe^y$ Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} = 0 + \left( x \cdot e^y \frac{dy}{dx} + e^y \cdot 1 \right)$ $\frac{dy}{dx} = xe^y \frac{dy}{dx} + e^y$ Rearranging the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} - xe^y \frac{dy}{dx} = e^y$ $\frac{dy}{dx} (1 - xe^y) = e^y$ $\frac{dy}{dx} = \frac{e^y}{1 - xe^y}$ From the original equation,we know $xe^y = y - 1$. Substituting this into the expression: $\frac{dy}{dx} = \frac{e^y}{1 - (y - 1)}$ $\frac{dy}{dx} = \frac{e^y}{1 - y + 1}$ $\frac{dy}{dx} = \frac{e^y}{2 - y}$

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 $y=1+xe^y$,then $\frac{dy}{dx}=$

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