The problem asks to determine $\frac{dy}{dx}$, $y'$, and $D_x y$ for three given functions. Let's analyze and solve them one by one:

a) $y = 2x + 5$

The derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(2x + 5) = 2$

Thus:

• $\frac{dy}{dx} = 2$
• $y' = 2$
• $D_x y = 2$

b) $y = x^2$

The derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x$

Thus:

• $\frac{dy}{dx} = 2x$
• $y' = 2x$
• $D_x y = 2x$

c) $y = \sqrt{x}$

Rewrite $\sqrt{x}$ as $x^{1/2}$, then differentiate: $\frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Thus:

• $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$
• $y' = \frac{1}{2\sqrt{x}}$
• $D_x y = \frac{1}{2\sqrt{x}}$

If you'd like, I can provide more detailed steps for each derivative or explain any specific part!

Related Questions:

1. What is the geometrical meaning of $\frac{dy}{dx}$?
2. How does the derivative change when the function becomes more complex (e.g., $y = x^3 + 2x + 1$)?
3. What is the relationship between higher-order derivatives and these first derivatives?
4. How can we interpret $D_x y$ in different notations?
5. How do we apply derivatives in practical scenarios, such as velocity and acceleration?

Tip:

When working with derivatives, always check if the function can be rewritten in a simpler form (like converting $\sqrt{x}$ to $x^{1/2}$) before differentiating. This can save time and reduce errors!