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

To find the derivatives of the given functions, we'll apply basic differentiation rules. (i) For \( y=3 \sqrt{x} \): Using the power rule, we rewrite \( \sqrt{x} \) as \( x^{1/2} \). Thus, \[ y' = \frac{d}{dx}(3x^{1/2}) = 3 \cdot \frac{1}{2} x^{-1/2} = \frac{3}{2\sqrt{x}}. \] (ii) For \( y=5x^{4}+4x-\frac{1}{2x}+\frac{1}{\sqrt{x}}-3 \): We can rewrite \( \frac{1}{2x} \) as \( \frac{1}{2} x^{-1} \) and \( \frac{1}{\sqrt{x}} \) as \( x^{-1/2} \). Now differentiating term by term gives us: \[ y' = 20x^{3} + 4 + \frac{1}{2} x^{-2} - \frac{1}{2} x^{-3/2} = 20x^{3} + 4 + \frac{1}{2x^{2}} - \frac{1}{2\sqrt{x^{3}}}. \] (iii) For \( y=\ln(4x)+\frac{5}{\sqrt{x^{3}}} \): First, we apply the rule for the derivative of a logarithm: \( \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx} \), where \( u = 4x \). Then, the derivative of \( \frac{5}{\sqrt{x^{3}}} \) can be rewritten as \( 5x^{-3/2} \): \[ y' = \frac{1}{4x} \cdot 4 + \frac{d}{dx}\left(5x^{-3/2}\right) = 1 - \frac{15}{2} x^{-5/2} = 1 - \frac{15}{2\sqrt{x^{5}}}. \] Thus, the derivatives of the functions are: (i) \( y' = \frac{3}{2\sqrt{x}} \) (ii) \( y' = 20x^{3} + 4 + \frac{1}{2x^{2}} - \frac{1}{2\sqrt{x^{3}}} \) (iii) \( y' = 1 - \frac{15}{2\sqrt{x^{5}}} \)