We are tasked with finding the second derivative of $y$ with respect to $x$, given the equation:

$\sqrt{y} + xy = 1$

Step 1: Differentiate the given equation

We begin by implicitly differentiating the equation with respect to $x$. Differentiating both sides of:

$\sqrt{y} + xy = 1$

with respect to $x$:

$\frac{d}{dx} \left( \sqrt{y} \right) + \frac{d}{dx} \left( xy \right) = \frac{d}{dx} (1)$

Derivative of $\sqrt{y}$:

Using the chain rule for $\sqrt{y} = y^{1/2}$, we get:

$\frac{d}{dx} \left( \sqrt{y} \right) = \frac{1}{2} y^{-1/2} \cdot \frac{dy}{dx}$

Derivative of $xy$:

Using the product rule, $\frac{d}{dx} (xy) = x \frac{dy}{dx} + y$.

So, the first derivative equation becomes:

$\frac{1}{2} y^{-1/2} \cdot \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$

Step 2: Solve for $\frac{dy}{dx}$

Now, solve for $\frac{dy}{dx}$:

$\left( \frac{1}{2} y^{-1/2} + x \right) \frac{dy}{dx} = -y$

$\frac{dy}{dx} = \frac{-y}{\frac{1}{2} y^{-1/2} + x}$

Step 3: Differentiate again to find $\frac{d^2y}{dx^2}$

Now, differentiate $\frac{dy}{dx}$ with respect to $x$. This requires the quotient rule. Let:

$u = -y \quad \text{and} \quad v = \frac{1}{2} y^{-1/2} + x$

Using the quotient rule:

$\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Derivative of $u = -y$:

$\frac{du}{dx} = -\frac{dy}{dx}$

Derivative of $v = \frac{1}{2} y^{-1/2} + x$:

$\frac{dv}{dx} = \frac{d}{dx} \left( \frac{1}{2} y^{-1/2} \right) + 1 = \frac{1}{2} \cdot \left( -\frac{1}{2} y^{-3/2} \right) \cdot \frac{dy}{dx} + 1 = -\frac{1}{4} y^{-3/2} \cdot \frac{dy}{dx} + 1$

Step 4: Substitute into the quotient rule

Now substitute $u$, $v$, $\frac{du}{dx}$, and $\frac{dv}{dx}$ into the quotient rule:

$\frac{d^2y}{dx^2} = \frac{\left( \frac{1}{2} y^{-1/2} + x \right) \left( -\frac{dy}{dx} \right) - (-y) \left( -\frac{1}{4} y^{-3/2} \cdot \frac{dy}{dx} + 1 \right)}{\left( \frac{1}{2} y^{-1/2} + x \right)^2}$

This is the expression for $\frac{d^2y}{dx^2}$. It can be simplified further, but the result involves both $y$ and $\frac{dy}{dx}$.

Would you like more details or have any questions on a specific part?

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Related Questions:

1. How do you solve implicit differentiation problems step by step?
2. What are the rules for applying the quotient rule in calculus?
3. How can the chain rule be applied in more complex derivatives?
4. How do second derivatives relate to concavity in a function?
5. How can you simplify expressions involving second derivatives?

Tip: When dealing with implicit differentiation, always carefully apply both the chain rule and product rule where necessary.