(ii) $$ y=5 x^{4}+4 x-\frac{1}{2 x}+\frac{1}{\sqrt{x}}-3 $$ (iii) $$ y=\ln 4 x+\frac{5}{\sqrt{x^{3}}} $$

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**Solution for (ii):**

We start with the function
$$
y = 5x^4 + 4x - \frac{1}{2x} + \frac{1}{\sqrt{x}} - 3.
$$
First, rewrite the terms with negative exponents:
- $$5x^4$$ stays the same.
- $$4x$$ stays the same.
- $$-\frac{1}{2x} = -\frac{1}{2} x^{-1}$$.
- $$\frac{1}{\sqrt{x}} = x^{-1/2}$$.
- The constant $$-3$$ has derivative 0.

Now, differentiate term by term using the power rule $$\frac{d}{dx}x^n = nx^{n-1}$$:

1. For $$5x^4$$:
   $$
   \frac{d}{dx}\left(5x^4\right) = 5 \cdot 4x^{4-1} = 20x^3.
   $$

2. For $$4x$$:
   $$
   \frac{d}{dx}(4x) = 4.
   $$

3. For $$-\frac{1}{2}x^{-1}$$:
   $$
   \frac{d}{dx}\left(-\frac{1}{2} x^{-1}\right) = -\frac{1}{2} \cdot (-1)x^{-1-1} = \frac{1}{2} x^{-2} = \frac{1}{2x^2}.
   $$

4. For $$x^{-1/2}$$:
   $$
   \frac{d}{dx}\left(x^{-1/2}\right) = -\frac{1}{2} x^{-1/2-1} = -\frac{1}{2} x^{-3/2} = -\frac{1}{2x^{3/2}}.
   $$

5. The derivative of the constant $$-3$$ is $$0$$.

Thus, combining all these results, we have:
$$
y' = 20x^3 + 4 + \frac{1}{2x^2} - \frac{1}{2x^{3/2}}.
$$

---

**Solution for (iii):**

We are given the function
$$
y = \ln(4x) + \frac{5}{\sqrt{x^3}}.
$$
It is assumed that $$\ln 4x$$ means $$\ln(4x)$$.

1. For $$\ln(4x)$$, notice that it can be split as:
   $$
   \ln(4x) = \ln 4 + \ln x.
   $$
   Since $$\ln 4$$ is constant, its derivative is 0. The derivative of $$\ln x$$ is:
   $$
   \frac{d}{dx} \ln x = \frac{1}{x}.
   $$
   Therefore,
   $$
   \frac{d}{dx}\left(\ln(4x)\right) = \frac{1}{x}.
   $$

2. For $$\frac{5}{\sqrt{x^3}}$$, first express it using exponents:
   $$
   \frac{5}{\sqrt{x^3}} = 5x^{-3/2}.
   $$
   Now, differentiate using the power rule:
   $$
   \frac{d}{dx}\left(5x^{-3/2}\right) = 5 \cdot \left(-\frac{3}{2}\right)x^{-3/2-1} = -\frac{15}{2}x^{-5/2}.
   $$

Thus, the derivative is:
$$
y' = \frac{1}{x} - \frac{15}{2}x^{-5/2}.
$$
**Derivatives:** **(ii)** $$ y' = 20x^3 + 4 + \frac{1}{2x^2} - \frac{1}{2x^{3/2}} $$ **(iii)** $$ y' = \frac{1}{x} - \frac{15}{2x^{5/2}} $$

(ii) \( y=5 x^{4}+4 x-\frac{1}{2 x}+\frac{1}{\sqrt{x}}-3 \) (iii) \( y=\ln 4 x+\frac{5}{\sqrt{x^{3}}} \)

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