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Part 1
Find StartFraction d squared y Over dx squared EndFraction
​, where StartRoot y EndRootplus2xyequals1.


just give me the exact formula for what the second derivative equals

Given the equation \(\sqrt{y} + 2xy = 1\), we need to find \(\frac{d^2 y}{dx^2}\).

First, let's rewrite the equation in a more convenient form:
\[ \sqrt{y} + 2xy = 1 \]

We differentiate both sides with respect to \(x\) to find \(\frac{dy}{dx}\):
\[ \frac{d}{dx}(\sqrt{y}) + \frac{d}{dx}(2xy) = \frac{d}{dx}(1) \]

Using the chain rule and product rule:
\[ \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} = 0 \]

Let's isolate \(\frac{dy}{dx}\):
\[ \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} + 2x \frac{dy}{dx} = -2y \]
\[ \left(\frac{1}{2\sqrt{y}} + 2x\right) \frac{dy}{dx} = -2y \]
\[ \frac{dy}{dx} = \frac{-2y}{\frac{1}{2\sqrt{y}} + 2x} \]

Now, let's simplify this:
\[ \frac{dy}{dx} = \frac{-2y}{\frac{1 + 4x\sqrt{y}}{2\sqrt{y}}} \]
\[ \frac{dy}{dx} = \frac{-2y \cdot 2\sqrt{y}}{1 + 4x\sqrt{y}} \]
\[ \frac{dy}{dx} = \frac{-4y\sqrt{y}}{1 + 4x\sqrt{y}} \]

Next, we differentiate \(\frac{dy}{dx}\) with respect to \(x\) to find \(\frac{d^2y}{dx^2}\):
\[ \frac{d}{dx} \left( \frac{-4y\sqrt{y}}{1 + 4x\sqrt{y}} \right) \]

Using the quotient rule:
\[ \frac{d^2 y}{dx^2} = \frac{(1 + 4x\sqrt{y}) \cdot \frac{d}{dx}(-4y\sqrt{y}) - (-4y\sqrt{y}) \cdot \frac{d}{dx}(1 + 4x\sqrt{y})}{(1 + 4x\sqrt{y})^2} \]

First, we find \(\frac{d}{dx}(-4y\sqrt{y})\):
\[ \frac{d}{dx}(-4y\sqrt{y}) = -4 \left( \frac{d}{dx}(y\sqrt{y}) \right) \]
Using the product rule and chain rule:
\[ \frac{d}{dx}(y\sqrt{y}) = \sqrt{y} \cdot \frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = \sqrt{y} \cdot \frac{dy}{dx} + \frac{y}{2\sqrt{y}} \cdot \frac{dy}{dx} = \left( \sqrt{y} + \frac{y}{2\sqrt{y}} \right) \frac{dy}{dx} = \frac{3}{2} \sqrt{y} \cdot \frac{dy}{dx} \]

So:
\[ \frac{d}{dx}(-4y\sqrt{y}) = -4 \cdot \frac{3}{2} \sqrt{y} \cdot \frac{dy}{dx} = -6 \sqrt{y} \cdot \frac{dy}{dx} \]

Next, we find \(\frac{d}{dx}(1 + 4x\sqrt{y})\):
\[ \frac{d}{dx}(1 + 4x\sqrt{y}) = 4 \left( \sqrt{y} + x \cdot \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \right) = 4\sqrt{y} + \frac{2x}{\sqrt{y}} \cdot \frac{dy}{dx} \]

Putting it all together:
\[ \frac{d^2 y}{dx^2} = \frac{(1 + 4x\sqrt{y})(-6 \sqrt{y} \cdot \frac{dy}{dx}) - (-4y\sqrt{y})(4\sqrt{y} + \frac{2x}{\sqrt{y}} \cdot \frac{dy}{dx})}{(1 + 4x\sqrt{y})^2} \]

Simplifying each term:
\[ = \frac{-6 \sqrt{y} \cdot \frac{dy}{dx} - 24x y \cdot \frac{dy}{dx} - (-16 y \cdot y - 8xy \cdot \frac{dy}{dx})}{(1 + 4x\sqrt{y})^2} \]
\[ = \frac{-6 \sqrt{y} \cdot \frac{dy}{dx} - 24xy \cdot \frac{dy}{dx} + 16y^2 + 8xy \cdot \frac{dy}{dx}}{(1 + 4x\sqrt{y})^2} \]
\[ = \frac{-6 \sqrt{y} \cdot \frac{dy}{dx} - 16xy \cdot \frac{dy}{dx} + 16y^2}{(1 + 4x\sqrt{y})^2} \]

Finally, substitute \(\frac{dy}{dx}\) back in to get the explicit formula for \(\frac{d^2 y}{dx^2}\).

Would you like further details or have any questions?

1. How do we apply the chain rule in differentiation?
2. What is the quotient rule and how is it used in differentiation?
3. How can implicit differentiation help solve complex equations?
4. What are some common applications of second derivatives?
5. How do product and chain rules differ in their applications?

**Tip:** Always simplify expressions step by step to avoid errors in differentiation.

Find the second derivative, \( \frac{d^2 y}{dx^2} \), where \( \sqrt{y} + 2xy = 1 \).

Explore how to find the second derivative \( \frac{d^2 y}{dx^2} \) of the equation \( \sqrt{y} + 2xy = 1 \) through implicit differentiation. This advanced problem involves using the chain rule, product rule, and quotient rule, making it suitable for students in advanced high school calculus or early college courses.

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