To find $\frac{dy}{dx}$ for the function $y = \sqrt{(x+1)(1+2x)}$, we will use the chain rule and the product rule for differentiation.

Let's go through the steps:

1. Rewrite the function for easier differentiation: $y = \sqrt{(x+1)(1+2x)} = ((x+1)(1+2x))^{\frac{1}{2}}$

2. Differentiate using the chain rule: Let $u = (x+1)(1+2x)$, so $y = u^{\frac{1}{2}}$.

   Then, $\frac{dy}{dx} = \frac{1}{2} u^{-\frac{1}{2}} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$

3. Differentiate $u = (x+1)(1+2x)$ using the product rule: $\frac{du}{dx} = (x+1) \cdot \frac{d}{dx}(1+2x) + (1+2x) \cdot \frac{d}{dx}(x+1)$

   Calculating each derivative:

   • $\frac{d}{dx}(1+2x) = 2$
   • $\frac{d}{dx}(x+1) = 1$

   So, $\frac{du}{dx} = (x+1) \cdot 2 + (1+2x) \cdot 1 = 2(x+1) + (1+2x)$ Simplify this: $\frac{du}{dx} = 2x + 2 + 1 + 2x = 4x + 3$

4. Substitute back into the expression for $\frac{dy}{dx}$: Since $u = (x+1)(1+2x)$, $\frac{dy}{dx} = \frac{1}{2\sqrt{(x+1)(1+2x)}} \cdot (4x + 3)$

Thus, the derivative is: $\frac{dy}{dx} = \frac{4x + 3}{2\sqrt{(x+1)(1+2x)}}$

Would you like further details or have any questions?

Here are some additional questions for deeper understanding:

1. How would you approach finding $\frac{d^2y}{dx^2}$ for this function?
2. What is the behavior of $\frac{dy}{dx}$ as $x \to \infty$?
3. How does the chain rule apply when there’s a nested square root in the function?
4. Can you identify any points where $\frac{dy}{dx}$ might be undefined?
5. How would you verify the correctness of this derivative using a different method?

Tip: Always simplify expressions inside the square root when possible, as it may reveal additional symmetries or allow for easier further differentiation.