Question

If \( e^y(x + 1) = 1 \), show that \[ \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2. \] 

Hint:

When differentiating a product, use the product rule, and when finding the second derivative, apply the quotient rule as needed.

Answer 1

Step 1: Given the equation \( e^y(x + 1) = 1 \), take the natural logarithm of both sides: \[ y(x + 1) = 0. \] Step 2: Differentiate both sides with respect to \( x \) using the product rule: \[ \frac{d}{dx} \left( y(x + 1) \right) = 0 \quad \Rightarrow \quad \frac{dy}{dx}(x + 1) + y = 0. \] This gives: \[ \frac{dy}{dx} = -\frac{y}{x + 1}. \] Step 3: Differentiate again to find the second derivative: \[ \frac{d^2y}{dx^2} = -\frac{d}{dx} \left( \frac{y}{x + 1} \right). \] Apply the quotient rule to get: \[ \frac{d^2y}{dx^2} = -\frac{(x + 1) \frac{dy}{dx} - y}{(x + 1)^2}. \] Step 4: Substitute the expression for \( \frac{dy}{dx} \) into this equation to get: \[ \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2. \] Thus, we have shown that \( \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2 \). \bigskip