From the first prinaiple, difterentiate with respect to \( x \) (i) \( \frac{-x}{2}+u \) (i) \( \frac{1}{2 \sqrt{x}} \)

Certainly! Let's differentiate each of the given functions with respect to \( x \) using the **first principles** (also known as the limit definition of the derivative). ### **First Principles of Differentiation** The derivative of a function \( f(x) \) at a point \( x \) is defined as: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] We'll apply this definition to both functions provided. --- ### **(i) Differentiating \( f(x) = \frac{-x}{2} + u \)** **Assumption:** Here, \( u \) is considered a constant with respect to \( x \). 1. **Apply the definition:** \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] 2. **Compute \( f(x+h) \):** \[ f(x+h) = \frac{-(x+h)}{2} + u \] 3. **Subtract \( f(x) \) from \( f(x+h) \):** \[ f(x+h) - f(x) = \left( \frac{-(x+h)}{2} + u \right) - \left( \frac{-x}{2} + u \right) = \frac{-h}{2} \] 4. **Form the difference quotient:** \[ \frac{f(x+h) - f(x)}{h} = \frac{-h/2}{h} = -\frac{1}{2} \] 5. **Take the limit as \( h \to 0 \):** \[ f'(x) = \lim_{h \to 0} \left( -\frac{1}{2} \right) = -\frac{1}{2} \] **Final Result:** \[ f'(x) = -\frac{1}{2} \] --- ### **(ii) Differentiating \( f(x) = \frac{1}{2 \sqrt{x}} \)** 1. **Rewrite the function for clarity:** \[ f(x) = \frac{1}{2} x^{-1/2} \] 2. **Apply the definition:** \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{2 \sqrt{x+h}} - \frac{1}{2 \sqrt{x}}}{h} \] 3. **Simplify the numerator:** \[ \frac{1}{2 \sqrt{x+h}} - \frac{1}{2 \sqrt{x}} = \frac{\sqrt{x} - \sqrt{x+h}}{2 \sqrt{x} \sqrt{x+h}} \] 4. **Form the difference quotient:** \[ \frac{\sqrt{x} - \sqrt{x+h}}{2 h \sqrt{x} \sqrt{x+h}} \] 5. **Rationalize the numerator:** Multiply numerator and denominator by \( \sqrt{x} + \sqrt{x+h} \): \[ \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{2 h \sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} = \frac{x - (x + h)}{2 h \sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} = \frac{-h}{2 h \sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} \] 6. **Simplify by canceling \( h \):** \[ \frac{-1}{2 \sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} \] 7. **Take the limit as \( h \to 0 \):** As \( h \to 0 \), \( \sqrt{x+h} \to \sqrt{x} \): \[ f'(x) = \frac{-1}{2 \cdot \sqrt{x} \cdot \sqrt{x} \cdot (2 \sqrt{x})} = \frac{-1}{4 x \sqrt{x}} = \frac{-1}{4 x^{3/2}} \] **Final Result:** \[ f'(x) = -\frac{1}{4 x^{3/2}} \] --- ### **Summary of Results** 1. **For \( f(x) = \frac{-x}{2} + u \):** \[ f'(x) = -\frac{1}{2} \] 2. **For \( f(x) = \frac{1}{2 \sqrt{x}} \):** \[ f'(x) = -\frac{1}{4 x^{3/2}} \]