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

**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: \[ y' = \frac{1}{x} - \frac{15}{2}x^{-5/2}. \]