2. Find \( \frac{d y}{d x} \) from first principles if \( y=\sqrt{1-2 x} \).

To find \( \frac{d y}{d x} \) from first principles, we start by using the definition of the derivative: \[ \frac{d y}{d x} = \lim_{h \to 0} \frac{y(x+h) - y(x)}{h}. \] We first compute \( y(x+h) \): \[ y(x+h) = \sqrt{1 - 2(x+h)} = \sqrt{1 - 2x - 2h}. \] Now, we substitute into the limit: \[ \frac{d y}{d x} = \lim_{h \to 0} \frac{\sqrt{1 - 2x - 2h} - \sqrt{1 - 2x}}{h}. \] To simplify this expression, we can multiply the numerator and the denominator by the conjugate: \[ \frac{d y}{d x} = \lim_{h \to 0} \frac{(\sqrt{1 - 2x - 2h} - \sqrt{1 - 2x})(\sqrt{1 - 2x - 2h} + \sqrt{1 - 2x})}{h(\sqrt{1 - 2x - 2h} + \sqrt{1 - 2x})}. \] This gives us: \[ = \lim_{h \to 0} \frac{ (1 - 2x - 2h) - (1 - 2x)}{h(\sqrt{1 - 2x - 2h} + \sqrt{1 - 2x})}. \] The numerator simplifies to: \[ -2h, \] thus: \[ \frac{d y}{d x} = \lim_{h \to 0} \frac{-2h}{h(\sqrt{1 - 2x - 2h} + \sqrt{1 - 2x})}. \] Canceling \( h \) gives: \[ \frac{d y}{d x} = \lim_{h \to 0} \frac{-2}{\sqrt{1 - 2x - 2h} + \sqrt{1 - 2x}}. \] Now, letting \( h \to 0 \): \[ \frac{d y}{d x} = \frac{-2}{\sqrt{1 - 2x} + \sqrt{1 - 2x}} = \frac{-2}{2\sqrt{1 - 2x}} = \frac{-1}{\sqrt{1 - 2x}}. \] So, the derivative is: \[ \frac{d y}{d x} = \frac{-1}{\sqrt{1 - 2x}}. \] --- Did you know that calculus, developed in the late 17th century, revolutionized mathematics by allowing for the precise study of motion and change? With contributions from geniuses like Isaac Newton and Gottfried Wilhelm Leibniz, this foundational tool helped explain everything from the orbits of planets to the behavior of light! If you're curious to dive deeper into calculus, exploring classic texts like "Calculus" by Michael Spivak can be insightful! Not only does it provide a solid foundation in theory, but it also includes rigorous problem sets that challenge your understanding and application of concepts in fun and engaging ways.