1. Use the definition of the derivative to find the derivative of the following functions: 1.1. Given $V(t) = \sqrt{14 + 3t}$. Using the definition, the derivative is $$V'(t) = \lim_{h \to 0} \frac{\sqrt{14 + 3(t+h)} - \sqrt{14 + 3t}}{h}.$$ Multiply numerator and denominator by the conjugate: $$ = \lim_{h \to 0} \frac{(\sqrt{14 + 3(t+h)} - \sqrt{14 + 3t})(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}{h(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}$$ $$ = \lim_{h \to 0} \frac{14 + 3(t+h) - (14 + 3t)}{h(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})} = \lim_{h \to 0} \frac{3h}{h(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}$$ $$= \lim_{h \to 0} \frac{3}{\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t}} = \frac{3}{2\sqrt{14+3t}}$$. 1.2. For $f(z) = z^2 + 3$, the derivative using the definition is $$f'(z) = \lim_{h \to 0} \frac{(z+h)^2 + 3 - (z^2 + 3)}{h} = \lim_{h \to 0} \frac{z^2 + 2zh + h^2 - z^2}{h} = \lim_{h \to 0} \frac{2zh + h^2}{h}$$ $$= \lim_{h \to 0} (2z + h) = 2z.$$ 1.3. For $W(t) = \frac{1}{\sqrt{t}} = t^{-\frac{1}{2}}$, the derivative using the definition is $$W'(t) = \lim_{h \to 0} \frac{(t+h)^{-\frac{1}{2}} - t^{-\frac{1}{2}}}{h}.$$ This simplifies using the binomial approximation or power rule to $$W'(t) = -\frac{1}{2} t^{-\frac{3}{2}} = -\frac{1}{2t^{3/2}}.$$ 2. Find the derivative of the given functions: 2.1. $W(x) = x^3 - \frac{1}{x^6} + \frac{1}{\sqrt[5]{x^2}} = x^3 - x^{-6} + x^{-\frac{2}{5}}$\ Using power rule: $$W'(x) = 3x^2 + 6x^{-7} - \frac{2}{5} x^{-\frac{7}{5}}.$$ 2.2. $S(w) = \frac{w^2(2-w) + w^5}{3w} = \frac{2w^2 - w^3 + w^5}{3w} = \frac{2w^2}{3w} - \frac{w^3}{3w} + \frac{w^5}{3w} = \frac{2w}{3} - \frac{w^2}{3} + \frac{w^4}{3}$ Taking derivative: $$S'(w) = \frac{2}{3} - \frac{2w}{3} + \frac{4w^3}{3}.$$ 2.3. $g(x) = (1 + 2x)(2 - x + x^2)$ Use product rule: $f'g + fg'$ $$f = 1 + 2x, f' = 2; \quad g = 2 - x + x^2, g' = -1 + 2x$$ So $$g'(x) = 2(2 - x + x^2) + (1 + 2x)(-1 + 2x)$$ $$= 4 - 2x + 2x^2 - 1 - 2x + 2x - 4x^2 = (4 - 1) + (-2x - 2x + 2x) + (2x^2 - 4x^2) = 3 - 2x - 2x^2.$$ 2.4. $g(x) = \frac{\sqrt[3]{x}}{1 + x^2} = \frac{x^{1/3}}{1 + x^2}$ Using quotient rule: $$g'(x) = \frac{(\frac{1}{3} x^{-2/3})(1 + x^2) - x^{1/3} (2x)}{(1 + x^2)^2}$$ Simplify numerator: $$= \frac{\frac{1}{3} x^{-2/3} (1 + x^2) - 2x^{4/3}}{(1 + x^2)^2}.$$ 2.5. $f(x) = \left(x - \frac{2}{x}\right)(7 - 2x^3)$ First rewrite: $$f(x) = (x - 2x^{-1})(7 - 2x^3).$$ Use product rule: $$f' = (1 + 2x^{-2})(7 - 2x^3) + (x - 2x^{-1})(-6x^2).$$ 2.6. $h(z) = (2 - \sqrt{z})(3 + 8 \sqrt[3]{z^2}) = (2 - z^{1/2})(3 + 8 z^{2/3})$ Use product rule: $$h'(z) = \left(-\frac{1}{2}z^{-1/2}\right)(3 + 8z^{2/3}) + (2 - z^{1/2}) \left(0 + 8 \cdot \frac{2}{3} z^{-1/3}\right).$$ 3. For $R(x) = (x + 1)(x - 2)^2$, find $R'(x)$: Use product rule: $$R'(x) = (1)(x - 2)^2 + (x + 1) \cdot 2(x - 2)$$ $$= (x - 2)^2 + 2(x + 1)(x - 2) = (x - 2)[(x - 2) + 2(x + 1)].$$ Simplify in bracket: $$(x - 2) + 2(x + 1) = x - 2 + 2x + 2 = 3x$$ So, $$R'(x) = (x - 2)(3x) = 3x(x - 2).$$ To find intervals where $R$ is increasing or decreasing, analyze sign of $R'(x)$: - Critical points at $x = 0$ and $x = 2$. - For $x < 0$: $3x < 0$, $x-2 < 0$; product positive. - For $0 < x < 2$: $3x > 0$, $x-2 < 0$; product negative. - For $x > 2$: both positive; product positive. So $R$ is increasing on $(-\infty, 0) \cup (2, \infty)$ and decreasing on $(0, 2)$. 4. For $f(x) = x^3 - 5x^2 + x$, find where tangent line is parallel to $y = 4x + 23$. Derivative: $$f'(x) = 3x^2 - 10x + 1.$$ Parallel means slope $= 4$: $$3x^2 - 10x + 1 = 4$$ $$3x^2 - 10x - 3 = 0$$ Solve quadratic: $$x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot (-3)}}{2\cdot 3} = \frac{10 \pm \sqrt{100 + 36}}{6} = \frac{10 \pm \sqrt{136}}{6}.$$ 5. For $f(x) = 8 + 4x + x^2 - 2x^3$, find where tangent line is perpendicular to $y = -\frac{1}{4}x + \frac{8}{3}$. Derivative: $$f'(x) = 4 + 2x - 6x^2.$$ The slope of given line is $-\frac{1}{4}$. For perpendicular lines, product of slopes is $-1$: $$f'(x) \cdot \left(-\frac{1}{4}\right) = -1 \Rightarrow f'(x) = 4.$$ Solve: $$4 + 2x - 6x^2 = 4$$ $$2x - 6x^2 = 0$$ $$2x(1 - 3x) = 0$$ So $$x = 0 \text{ or } x = \frac{1}{3}.$$