91Ó°ÊÓ

Differentiate both sides of the equation implicitly with respect to x. Differentiate each term: Given equation: \( xy + x - 2y = 4 \) Apply the product rule to \( xy \), and remember to differentiate \( y \) implicitly: \( \frac{d}{dx}(xy) + \frac{d}{dx}(x) - \frac{d}{dx}(2y) = \frac{d}{dx}(4) \) Using the product rule: \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \). Differentiating the rest: \( x \frac{dy}{dx} + y + 1 - 2 \frac{dy}{dx} = 0 \)

Rearrange the equation to isolate \( \frac{dy}{dx} \): \( x \frac{dy}{dx} + y + 1 - 2 \frac{dy}{dx} = 0 \) Combine like terms involving \( \frac{dy}{dx} \): \( x \frac{dy}{dx} - 2 \frac{dy}{dx} = -y - 1 \) Simplify further: \( (x - 2) \frac{dy}{dx} = -y - 1 \) Finally: \( \frac{dy}{dx} = \frac{-y - 1}{x - 2} \)

Now, differentiate \( \frac{dy}{dx} = \frac{-y - 1}{x - 2} \) implicitly to find \( \frac{d^2y}{dx^2} \). Use the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Let: \( u = -y - 1 \) \( v = x - 2 \) Find derivatives: \( u' = -\frac{dy}{dx} \) and \( v' = 1 \) Substituting values: \( \frac{d}{dx} \left( \frac{-y - 1}{x - 2} \right) = \frac{(-\frac{dy}{dx})(x - 2) - (-y - 1)(1)}{(x - 2)^2} \) Simplify the numerator: \( = \frac{-\frac{dy}{dx}(x - 2) + y + 1}{(x - 2)^2} \). Substitute \( \frac{dy}{dx} = \frac{-y - 1}{x - 2} \): \( = \frac{-\left( \frac{-y - 1}{x - 2} \right)(x - 2) + y + 1}{(x - 2)^2} \) Simplify: \( = \frac{y + 1 + y + 1}{(x - 2)^2} \) Finally: \( \frac{d^2y}{dx^2} = \frac{2(y + 1)}{(x - 2)^2} \)