Question1. Use the definition to find the derivative of each of the following functions: (a) $f(x):=x^{3}$ for $x \in \mathbb{R}$, (b) $g(x):=1 / x$ for $x \in \mathbb{R}, x \neq 0$, (c) $h(x):=\sqrt{x}$ for $x>0$, (d) $k(x):=1 / \sqrt{x}$ for $x>0$.

Studdy Solution
Combine the fractions in the numerator:
$$k'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}$$
Multiply the numerator and the denominator by the conjugate:
$$k'(x) = \lim_{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$$
$$k'(x) = \lim_{h \to 0} \frac{x - (x+h)}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$$
$$k'(x) = \lim_{h \to 0} \frac{-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$$
Cancel $h$ in the numerator and denominator:
$$k'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$$
Evaluate the limit as $h \to 0$:
$$k'(x) = -\frac{1}{2x\sqrt{x}}$$
The derivatives of the functions are: (a) $f'(x) = 3x^2$ (b) $g'(x) = -\frac{1}{x^2}$ (c) $h'(x) = \frac{1}{2\sqrt{x}}$ (d) $k'(x) = -\frac{1}{2x\sqrt{x}}$