QUESTION ONE ( 16 MARKS) (a) Compute the derivative of \( y=\frac{1}{x} \) using the first principle (or by definition).

To find the derivative of \( y = \frac{1}{x} \) using the first principle, we start with the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substituting \( f(x) = \frac{1}{x} \): \[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \] Next, we need to combine the fractions in the numerator: \[ f'(x) = \lim_{h \to 0} \frac{\frac{x - (x + h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)} \] Here, the \( h \) in the numerator and denominator cancels out (as long as \( h \neq 0 \)): \[ f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} \] As \( h \) approaches 0, the expression simplifies to: \[ f'(x) = \frac{-1}{x^2} \] Thus, the derivative of \( y = \frac{1}{x} \) is: \[ f'(x) = -\frac{1}{x^2} \] --- The calculus concept of derivatives dates back to the 17th century with mathematicians like Newton and Leibniz developing foundational ideas that would shape modern calculus. The principles they established revolutionized our understanding of motion, area under curves, and rates of change, laying the groundwork for a wide array of scientific applications from physics to economics. In the real world, understanding derivatives is essential in various fields such as physics, where they describe velocity and acceleration, or in economics, where derivatives can indicate marginal cost and revenue. For instance, knowing how to calculate the derivative of a cost function allows businesses to make informed decisions about pricing, production levels, and overall profitability. By applying these concepts, professionals can effectively strategize for growth and efficiency!