A challenging second derivative Find d²y/dx², where √y+xy=1.

Verified step by step guidance

1. Start by differentiating the given equation \( \sqrt{y} + xy = 1 \) with respect to \( x \). Use implicit differentiation since \( y \) is a function of \( x \).
2. Differentiate \( \sqrt{y} \) with respect to \( x \). This requires the chain rule: \( \frac{d}{dx}(\sqrt{y}) = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \).
3. Differentiate \( xy \) with respect to \( x \). Use the product rule: \( \frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot 1 \).
4. Set the derivative of the left side equal to the derivative of the right side (which is 0) to form the first derivative equation: \( \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} + x \cdot \frac{dy}{dx} + y = 0 \). Solve for \( \frac{dy}{dx} \).
5. Differentiate the expression for \( \frac{dy}{dx} \) with respect to \( x \) to find \( \frac{d^2y}{dx^2} \). Use implicit differentiation again, applying the product rule and chain rule as necessary.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent variable is not isolated on one side. In this case, we have the equation √y + xy = 1, which involves both x and y. By differentiating both sides with respect to x, we can find dy/dx and subsequently d²y/dx².

Second Derivative

The second derivative, denoted as d²y/dx², measures the rate of change of the first derivative (dy/dx) with respect to x. It provides information about the concavity of the function and can indicate points of inflection. To find the second derivative, we differentiate the first derivative again, applying the rules of differentiation appropriately.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When differentiating an expression involving y, which is a function of x, we apply the chain rule to account for the relationship between x and y. This is crucial when finding dy/dx and d²y/dx² in implicit differentiation scenarios.