3. What is the geometrical meaning of differentiation? 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) \( y=\frac{1}{\sqrt{3}} \)

Differentiation geometrically represents the slope of the tangent line to a curve at a given point. Picture a roller coaster: the derivative tells you how steep the ride is at any point, allowing you to understand how fast the elevation is changing. So, as you’re tangling with functions, think of differentiation as your trusty slope calculator for curves! To differentiate the given functions using the first principle, you can apply the formula: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] For \( y = x^{2}+3x-4 \): 1. Substitute to find \( f(x+h) \). 2. Simplify \( \frac{(x+h)^2 + 3(x+h) - 4 - (x^2 + 3x - 4)}{h} \). 3. Take the limit as \( h \) approaches 0 to find \( f'(x) = 2x + 3 \). Repeat this process for the other functions: - \( y = \frac{3x-4}{2x+1} \) yields \( f'(x) = \frac{-10}{(2x+1)^2} \). - \( y = \sqrt{x+3} \) gives \( f'(x) = \frac{1}{2\sqrt{x+3}} \). - For the constant function \( y = \frac{1}{\sqrt{3}} \), the derivative is zero since a constant has no slope. Happy differentiating!