1. Use the definition of the derivative to find the derivatives. 1.a. Given $Q(t)=10+5t-t^2$, the definition of derivative is $$Q'(t)=\lim_{h\to0}\frac{Q(t+h)-Q(t)}{h}.$$ Calculate $Q(t+h) = 10 + 5(t+h) - (t+h)^2 = 10 + 5t + 5h - t^2 - 2th - h^2.$ Then, $$\frac{Q(t+h)-Q(t)}{h}=\frac{10+5t+5h -t^2 - 2th - h^2 - (10+5t - t^2)}{h} = \frac{5h - 2th - h^2}{h}=5 - 2t - h.$$ Taking the limit as $h \to 0$, $$Q'(t) = 5 - 2t.$$ \boxed{Q'(t)=5-2t}. 1.b. For $g(x)=x^3 - 2x^2 + x - 1$, $$g'(x)=\lim_{h\to0}\frac{(x+h)^3 -2(x+h)^2 + (x+h) -1 - (x^3 - 2x^2 + x - 1)}{h}.$$ Expand: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ $(x+h)^2 = x^2 + 2xh + h^2$ Substitute: $$\frac{x^3 + 3x^2h + 3xh^2 + h^3 - 2(x^2 + 2xh + h^2) + x + h -1 - x^3 + 2x^2 - x + 1}{h} = $$ $$\frac{3x^2h + 3xh^2 + h^3 - 2x^2 - 4xh - 2h^2 + x + h -1 + 2x^2 - x + 1 }{h} = \frac{3x^2h + 3xh^2 + h^3 -4xh - 2h^2 + h}{h}.$$ Divide by $h$: $$3x^2 + 3xh + h^2 - 4x - 2h + 1.$$ Taking limit $h \to 0$, $$g'(x) = 3x^2 - 4x + 1.$$ \boxed{g'(x) = 3x^2 - 4x + 1}. 1.c. For $V(t) = \frac{t+1}{t+4}$, $$V'(t)=\lim_{h\to0}\frac{\frac{t+h+1}{t+h+4} - \frac{t+1}{t+4}}{h}.$$ Combine fraction: $$\frac{(t+h+1)(t+4) - (t+1)(t+h+4)}{h(t+h+4)(t+4)}.$$ Expand numerator: $(t+h+1)(t+4) = t^2 + th + 4t + 4h + t + h + 4$ $(t+1)(t+h+4) = t^2 + th + 4t + t + h + 4$ Subtract: $$[t^2 + th + 4t + 4h + t + h + 4] - [t^2 + th +4t + t + h + 4] = (4h + h) - (0) = 5h.$$ So quotient is $$\frac{5h}{h(t+h+4)(t+4)} = \frac{5}{(t+h+4)(t+4)}.$$ Take limit $h \to 0$, $$V'(t) = \frac{5}{(t+4)^2}.$$ \boxed{V'(t) = \frac{5}{(t+4)^2}}. 1.d. For $Z(t) = \sqrt{3t - 4} = (3t - 4)^{1/2}$, $$Z'(t) = \lim_{h\to0} \frac{\sqrt{3(t+h)-4} - \sqrt{3t -4}}{h} = \lim_{h\to0} \frac{\sqrt{3t+3h-4} - \sqrt{3t -4}}{h}.$$ Multiply numerator and denominator by conjugate: $$\frac{[\sqrt{3t+3h-4} - \sqrt{3t -4}][\sqrt{3t+3h-4}+ \sqrt{3t -4}]}{h[\sqrt{3t+3h-4}+ \sqrt{3t -4}]} = \frac{3h}{h[\sqrt{3t+3h-4}+ \sqrt{3t -4}]} = \frac{3}{\sqrt{3t+3h-4}+ \sqrt{3t -4}}.$$ Taking limit $h \to 0$: $$Z'(t) = \frac{3}{2 \sqrt{3t - 4}}.$$ \boxed{Z'(t) = \frac{3}{2 \sqrt{3t - 4}}}. 2. Find derivatives using formulas. 2.a. $f(x) = 10 \sqrt[4]{x^3} - \sqrt{x^7} + 6 \sqrt[3]{x^8} - 3$. Rewrite with exponents: $$10 x^{3/4} - x^{7/2} + 6 x^{8/3} - 3.$$ Apply power rule: $$f'(x) = 10 \cdot \frac{3}{4} x^{3/4 -1} - \frac{7}{2} x^{7/2 -1} + 6 \cdot \frac{8}{3} x^{8/3 -1} - 0 = \frac{30}{4} x^{-1/4} - \frac{7}{2} x^{5/2} + 16 x^{5/3}.$$ Simplify fraction $\frac{30}{4} = \frac{15}{2}$: \boxed{f'(x)= \frac{15}{2} x^{-1/4} - \frac{7}{2} x^{5/2} + 16 x^{5/3}}. 2.b. $y = \left(1 + \sqrt{x^3}\right) \left(x^{-3} - 2 \sqrt[3]{x}\right)$. Rewrite: $$y = (1 + x^{3/2})(x^{-3} - 2 x^{1/3}).$$ Use product rule $y' = u' v + u v'$, where $u=1+x^{3/2}$, $u' = \frac{3}{2} x^{1/2}$, $v = x^{-3} - 2 x^{1/3}$, $v' = -3 x^{-4} - \frac{2}{3} x^{-2/3}$. Then $$y' = \frac{3}{2} x^{1/2} (x^{-3} - 2 x^{1/3}) + (1 + x^{3/2})(-3 x^{-4} - \frac{2}{3} x^{-2/3}).$$ Simplify each term or leave as is for clarity. \boxed{y' = \frac{3}{2} x^{1/2} (x^{-3} - 2 x^{1/3}) + (1 + x^{3/2})(-3 x^{-4} - \frac{2}{3} x^{-2/3})}. 2.c. $h(y) = \frac{y}{1 - e^y}$. Use quotient rule: $$h'(y) = \frac{(1 - e^y)(1) - y(- e^y)}{(1 - e^y)^2} = \frac{1 - e^y + y e^y}{(1 - e^y)^2}.$$ Simplify numerator: $$1 - e^y + y e^y = 1 + e^y (y - 1).$$ \boxed{h'(y) = \frac{1 + e^y (y - 1)}{(1 - e^y)^2}}. 2.d. $f(w) = \tan(w) \sec(w)$. Use product rule: $$f'(w) = \sec^2(w) \sec(w) + \tan(w) (\sec(w) \tan(w)) = \sec^3(w) + \tan^2(w) \sec(w).$$ Factor $\sec(w)$: $$f'(w) = \sec(w)(\sec^2(w) + \tan^2(w)).$$ Using identity $\sec^2(w) = 1 + \tan^2(w)$: $$\sec^2(w) + \tan^2(w) = (1 + \tan^2(w)) + \tan^2(w) = 1 + 2 \tan^2(w).$$ So \boxed{f'(w) = \sec(w)(1 + 2 \tan^2(w))}. 2.e. $f(x) = \sinh(x) + 2 \cosh(x) - \operatorname{sech}(x)$. Derivative: $$f'(x) = \cosh(x) + 2 \sinh(x) - (-\operatorname{sech}(x) \tanh(x)) = \cosh(x) + 2 \sinh(x) + \operatorname{sech}(x) \tanh(x).$$ \boxed{f'(x) = \cosh(x) + 2 \sinh(x) + \operatorname{sech}(x) \tanh(x)}. 2.f. $g(x) = \left(\ln(x^2 + 1) - \tan^{-1}(6x)\right)^{10}$ Use chain rule: Let $u = \ln(x^2 +1) - \tan^{-1}(6x)$, then $$g'(x) = 10 u^9 \cdot u'.$$ Calculate $u'$: $$u' = \frac{2x}{x^2 +1} - \frac{6}{1 + (6x)^2} = \frac{2x}{x^2 +1} - \frac{6}{1 + 36 x^2}.$$ Hence, \boxed{g'(x) = 10 \left(\ln(x^2 +1) - \tan^{-1}(6x)\right)^9 \left(\frac{2x}{x^2 +1} - \frac{6}{1 + 36 x^2}\right)}.