Let y = frac{1}{1 + x + ln x}. Then, | vedclass

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(B) Given $y = \frac{1}{1 + x + \ln x}$.Taking the reciprocal, we get $\frac{1}{y} = 1 + x + \ln x$.Differentiating both sides with respect to $x$:$-\

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$x \frac{dy}{dx} = y(y \ln x - 1)$

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(B) Given $y = \frac{1}{1 + x + \ln x}$.Taking the reciprocal, we get $\frac{1}{y} = 1 + x + \ln x$.Differentiating both sides with respect to $x$:$-\frac{1}{y^{2}} \frac{dy}{dx} = 1 + \frac{1}{x} = \frac{x + 1}{x}$.From the original equation, $1 + \ln x = \frac{1}{y} - x$.Substitute this into the expression for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{y^{2}(x + 1)}{x}$.Alternatively, differentiating $y(1 + x + \ln x) = 1$ with respect to $x$:$\frac{dy}{dx}(1 + x + \ln x) + y(1 + \frac{1}{x}) = 0$.Since $1 + x + \ln x = \frac{1}{y}$, we have:$\frac{dy}{dx} \cdot \frac{1}{y} + y \left(\frac{x + 1}{x}\right) = 0$.$\frac{1}{y} \frac{dy}{dx} = -y \left(\frac{x + 1}{x}\right)$.$x \frac{dy}{dx} = -y^{2}(x + 1)$.Wait, let us re-evaluate: $\frac{1}{y} = 1 + x + \ln x \implies \frac{d}{dx}(\frac{1}{y}) = \frac{d}{dx}(1 + x + \ln x) \implies -\frac{1}{y^{2}} \frac{dy}{dx} = 1 + \frac{1}{x} = \frac{x + 1}{x}$.$\frac{dy}{dx} = -\frac{y^{2}(x + 1)}{x}$.Checking option $B$: $x \frac{dy}{dx} = y(y \ln x - 1)$.If $y = \frac{1}{1 + x + \ln x}$, then $y \ln x - 1 = y \ln x - y(1 + x + \ln x) = y(\ln x - 1 - x - \ln x) = y(-1 - x) = -y(1 + x)$.So $y(y \ln x - 1) = y(-y(1 + x)) = -y^{2}(1 + x)$.Thus, $x \frac{dy}{dx} = -y^{2}(1 + x)$, which matches.

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

Let $y = \frac{1}{1 + x + \ln x}$. Then,

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