Questions: Evaluate d^2 y / d x^2 at the point (3,2). x y - x + y = 5

Solution Steps

To find the second derivative \( \frac{d^2 y}{dx^2} \) at the point \( (3, 2) \) for the equation \( xy - x + y = 5 \), we need to follow these steps:

1. Differentiate the given equation implicitly with respect to \( x \) to find \( \frac{dy}{dx} \).
2. Differentiate the result again with respect to \( x \) to find \( \frac{d^2 y}{dx^2} \).
3. Substitute the point \( (3, 2) \) into the expressions for \( \frac{dy}{dx} \) and \( \frac{d^2 y}{dx^2} \) to evaluate the second derivative at that point.

To solve the problem of evaluating the second derivative \( \frac{d^2 y}{dx^2} \) at the point \((3, 2)\) for the equation \( xy - x + y = 5 \), we will follow these steps:

Given the equation: \[ xy - x + y = 5 \]

Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(xy) - \frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dx}(5) \]

Using the product rule for the term \(xy\): \[ \frac{d}{dx}(xy) = x\frac{dy}{dx} + y \]

Thus, the differentiation becomes: \[ x\frac{dy}{dx} + y - 1 + \frac{dy}{dx} = 0 \]

Combine like terms: \[ (x + 1)\frac{dy}{dx} + y - 1 = 0 \]

Solve for \(\frac{dy}{dx}\): \[ (x + 1)\frac{dy}{dx} = 1 - y \] \[ \frac{dy}{dx} = \frac{1 - y}{x + 1} \]

Differentiate \(\frac{dy}{dx} = \frac{1 - y}{x + 1}\) with respect to \(x\): \[ \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{1 - y}{x + 1}\right) \]

Using the quotient rule: \[ \frac{d}{dx}\left(\frac{1 - y}{x + 1}\right) = \frac{(x + 1)\frac{d}{dx}(1 - y) - (1 - y)\frac{d}{dx}(x + 1)}{(x + 1)^2} \]

Calculate each derivative: \[ \frac{d}{dx}(1 - y) = -\frac{dy}{dx} \] \[ \frac{d}{dx}(x + 1) = 1 \]

Substitute these into the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{(x + 1)(-\frac{dy}{dx}) - (1 - y)(1)}{(x + 1)^2} \]

Substitute \(\frac{dy}{dx} = \frac{1 - y}{x + 1}\): \[ \frac{d^2y}{dx^2} = \frac{-(x + 1)\left(\frac{1 - y}{x + 1}\right) - (1 - y)}{(x + 1)^2} \] \[ = \frac{-(1 - y) - (1 - y)}{(x + 1)^2} \] \[ = \frac{-2(1 - y)}{(x + 1)^2} \]

Substitute \(x = 3\) and \(y = 2\) into the expression for \(\frac{d^2y}{dx^2}\): \[ \frac{d^2y}{dx^2} = \frac{-2(1 - 2)}{(3 + 1)^2} \] \[ = \frac{-2(-1)}{4^2} \] \[ = \frac{2}{16} \] \[ = \frac{1}{8} \]

Final Answer