The task asks to find the derivative of the function $y = \sqrt{2x + 1}$ using the general differentiation rule.

Solution:

The function can be rewritten as: $y = (2x + 1)^{\frac{1}{2}}.$

To differentiate, apply the chain rule:

1. Differentiate the outer function $u^{\frac{1}{2}}$ with respect to $u$:
   $\frac{d}{du}\left(u^{\frac{1}{2}}\right) = \frac{1}{2}u^{-\frac{1}{2}}.$

2. Differentiate the inner function $2x + 1$ with respect to $x$:
   $\frac{d}{dx}(2x + 1) = 2.$

Combine the results using the chain rule: $\frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-\frac{1}{2}} \cdot 2.$

Simplify: $\frac{dy}{dx} = \frac{2}{2} \cdot (2x + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x + 1}}.$

Final Answer:

The derivative of $y = \sqrt{2x + 1}$ is: $\frac{dy}{dx} = \frac{1}{\sqrt{2x + 1}}.$

Would you like further clarification or a breakdown of the steps?

Related Questions:

1. What is the derivative of $y = \sqrt{ax + b}$?
2. How does the chain rule work in differentiation?
3. What is the physical interpretation of the derivative of $\sqrt{2x+1}$?
4. How can we simplify expressions involving derivatives of square roots?
5. What is the second derivative of $y = \sqrt{2x+1}$?

Tip:

When differentiating functions with nested compositions (like square roots or exponents), always identify and separately differentiate the outer and inner functions using the chain rule.