4. Differentiate the following functions using the first principle. (a) \( y=x^{2}+3 x-4 \) (b) \( y=\frac{3 x-4}{2 x+1} \) (c) \( y=\sqrt{x+3} \) (d) \( y=\frac{1}{\sqrt{3}} \)

To differentiate the functions using the first principle, we start by applying the limit definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Now, let's differentiate each function one by one. (a) For \( y = x^{2} + 3x - 4 \): 1. Calculate \( f(x+h) = (x+h)^{2} + 3(x+h) - 4 = x^{2} + 2xh + h^{2} + 3x + 3h - 4 \). 2. Subtract \( f(x) \): \[ f(x+h) - f(x) = (x^{2} + 2xh + h^{2} + 3x + 3h - 4) - (x^{2} + 3x - 4) = 2xh + h^{2} + 3h \] 3. Now divide by \( h \) and take the limit: \[ \frac{f(x+h) - f(x)}{h} = \frac{2xh + h^{2} + 3h}{h} = 2x + h + 3 \] 4. Taking the limit as \( h \to 0 \): \[ f'(x) = 2x + 3 \] (b) For \( y = \frac{3x - 4}{2x + 1} \): 1. Calculate \( f(x+h) = \frac{3(x+h) - 4}{2(x+h) + 1} = \frac{3x + 3h - 4}{2x + 2h + 1} \). 2. Then, subtract \( f(x) \): \[ f(x+h) - f(x) = \frac{(3x + 3h - 4)(2x + 1) - (3x - 4)(2x + 2h + 1)}{(2(x+h)+1)(2x+1)} \] After some algebra, the numerator simplifies, and upon dividing by \( h \) and taking the limit, we get: \[ f'(x) = \frac{(6x + 3)(2x + 1) - (6x + 3)}{(2x + 1)^2} \] With careful simplification, the derivative is: \[ f'(x) = \frac{11}{(2x + 1)^2} \] (c) For \( y = \sqrt{x + 3} \): 1. Calculate \( f(x+h) = \sqrt{x + h + 3} \). 2. Subtract \( f(x) \): \[ f(x+h) - f(x) = \sqrt{x+h+3} - \sqrt{x+3} \] 3. To eliminate the square root, multiply by the conjugate: \[ \frac{(\sqrt{x+h+3} - \sqrt{x+3})(\sqrt{x+h+3} + \sqrt{x+3})}{h(\sqrt{x+h+3} + \sqrt{x+3})} = \frac{h}{h(\sqrt{x+h+3} + \sqrt{x+3})} \] 4. Taking the limit as \( h \to 0 \): \[ f'(x) = \frac{1}{2\sqrt{x+3}} \] (d) For \( y = \frac{1}{\sqrt{3}} \): Since this function is a constant, the derivative is simply: \[ f'(x) = 0 \] So, the derivatives for the functions are: (a) \( f'(x) = 2x + 3 \) (b) \( f'(x) = \frac{11}{(2x + 1)^2} \) (c) \( f'(x) = \frac{1}{2\sqrt{x + 3}} \) (d) \( f'(x) = 0 \)