Let's solve these differentiation problems one by one:

(a) $y = x^2$ $\frac{dy}{dx} = 2x$

(b) $y = x^5$ $\frac{dy}{dx} = 5x^4$

(c) $y = x^{-1}$ $\frac{dy}{dx} = -x^{-2} = -\frac{1}{x^2}$

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

(e) $y = \pi$ \frac{dy}{dx} = 0 \quad \text{(since \pi is a constant)}

(f) $y = 3x^4$ $\frac{dy}{dx} = 3 \cdot 4x^3 = 12x^3$

(g) $y = x^2 + 3$ $\frac{dy}{dx} = 2x + 0 = 2x$

(h) $y = x^5 - 6x^7$ $\frac{dy}{dx} = 5x^4 - 6 \cdot 7x^6 = 5x^4 - 42x^6$

(i) $y = \frac{2x^2 - 3}{\sqrt{x}}$

First, rewrite $y$ as: $y = \frac{2x^2 - 3}{x^{1/2}} = 2x^{3/2} - 3x^{-1/2}$ Now, differentiate: $\frac{dy}{dx} = 2 \cdot \frac{3}{2}x^{1/2} - 3 \cdot \left(-\frac{1}{2}\right)x^{-3/2}$ $\frac{dy}{dx} = 3x^{1/2} + \frac{3}{2}x^{-3/2}$

(j) $y = (x + 1)(x^2 + 3)$

Use the product rule: $\frac{dy}{dx} = (x+1) \frac{d}{dx}(x^2+3) + (x^2+3) \frac{d}{dx}(x+1)$ $= (x+1)(2x) + (x^2+3)(1)$ $= 2x(x+1) + (x^2 + 3) = 2x^2 + 2x + x^2 + 3$ $\frac{dy}{dx} = 3x^2 + 2x + 3$

(k) $y = \frac{3x^2 - 1}{x}$

First, rewrite $y$ as: $y = \frac{3x^2}{x} - \frac{1}{x} = 3x - x^{-1}$ Now, differentiate: $\frac{dy}{dx} = 3 + x^{-2} = 3 + \frac{1}{x^2}$

(l) $y = \sqrt{x(x+1)}$

First, rewrite $y$ as: $y = (x(x+1))^{1/2} = (x^2 + x)^{1/2}$ Now, apply the chain rule: $\frac{dy}{dx} = \frac{1}{2}(x^2 + x)^{-1/2} \cdot (2x + 1)$ $\frac{dy}{dx} = \frac{2x+1}{2\sqrt{x^2 + x}}$

Would you like more details on any of these steps or have any questions?

Here are some related problems to consider:

1. Differentiate $y = x^3$.
2. Differentiate $y = \sin(x)$.
3. Differentiate $y = e^x$.
4. Differentiate $y = \ln(x)$.
5. Differentiate $y = \frac{1}{x+2}$.

Tip: Remember to always simplify functions before differentiating, if possible!