1. Problem 108: Find $f'(x)$ for $f(x) = (x^2 + 1)(x^3 + 3)$ in two ways. (a) Multiply first, then differentiate: $$f(x) = (x^2 + 1)(x^3 + 3) = x^2 \cdot x^3 + x^2 \cdot 3 + 1 \cdot x^3 + 1 \cdot 3 = x^5 + 3x^2 + x^3 + 3$$ Simplify: $$f(x) = x^5 + x^3 + 3x^2 + 3$$ Differentiate term-by-term: $$f'(x) = 5x^4 + 3x^2 + 6x + 0 = 5x^4 + 3x^2 + 6x$$ (b) Use the product rule: Recall product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ Let $u = x^2 + 1$, $v = x^3 + 3$ Compute derivatives: $$u' = 2x, \quad v' = 3x^2$$ Apply product rule: $$f'(x) = u'v + uv' = 2x(x^3 + 3) + (x^2 + 1)(3x^2) = 2x^4 + 6x + 3x^4 + 3x^2 = (2x^4 + 3x^4) + 3x^2 + 6x = 5x^4 + 3x^2 + 6x$$ Both methods give the same derivative: $$f'(x) = 5x^4 + 3x^2 + 6x$$ 2. Problem 109: Find $f'(x)$ for $f(x) = (1 + x^2)^4$ in two ways. (a) Expand and differentiate: Use binomial expansion: $$(1 + x^2)^4 = \sum_{k=0}^4 \binom{4}{k} (1)^{4-k} (x^2)^k = 1 + 4x^2 + 6x^4 + 4x^6 + x^8$$ Differentiate term-by-term: $$f'(x) = 0 + 8x + 24x^3 + 24x^5 + 8x^7$$ (b) Use power rule and chain rule: $$f(x) = (1 + x^2)^4$$ Let $g(x) = 1 + x^2$, then $f(x) = g(x)^4$ By chain rule: $$f'(x) = 4g(x)^3 \cdot g'(x) = 4(1 + x^2)^3 \cdot 2x = 8x(1 + x^2)^3$$ Check if this matches the expanded derivative by expanding $(1 + x^2)^3$: $$(1 + x^2)^3 = 1 + 3x^2 + 3x^4 + x^6$$ Multiply by $8x$: $$8x + 24x^3 + 24x^5 + 8x^7$$ This matches the expanded derivative exactly. 3. Problem 110: Prove $(cu)' = c(u')$ from the product rule. Let $c$ be a constant and $u = u(x)$ a differentiable function. By product rule: $$\frac{d}{dx}[cu] = c \cdot \frac{d}{dx}[u] + u \cdot \frac{d}{dx}[c]$$ Since $c$ is constant, $\frac{d}{dx}[c] = 0$, so: $$\frac{d}{dx}[cu] = c u' + u \cdot 0 = c u'$$ Hence proved. 4. Problem 111: $f(x) = x + 1 + (x + 1)^2$ Differentiate term-by-term: $$f'(x) = 1 + 0 + 2(x + 1) \cdot 1 = 1 + 2(x + 1) = 1 + 2x + 2 = 2x + 3$$ 5. Problem 112: $f(x) = \frac{x - 2}{x^4 + 1}$ Use quotient rule: $$f'(x) = \frac{(x^4 + 1)(1) - (x - 2)(4x^3)}{(x^4 + 1)^2} = \frac{x^4 + 1 - 4x^3(x - 2)}{(x^4 + 1)^2}$$ Simplify numerator: $$x^4 + 1 - 4x^4 + 8x^3 = -3x^4 + 8x^3 + 1$$ So: $$f'(x) = \frac{-3x^4 + 8x^3 + 1}{(x^4 + 1)^2}$$ 6. Problem 113: $f(x) = \left(\frac{1}{1 + x}\right)^{-1}$ Rewrite: $$f(x) = (1 + x)^1 = 1 + x$$ Differentiate: $$f'(x) = 1$$ 7. Problem 114: $f(x) = \sqrt{1 - x^2} = (1 - x^2)^{1/2}$ Use chain rule: $$f'(x) = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}}$$ 8. Problem 115: $f(x) = \frac{ax + b}{cx + d}$ Use quotient rule: $$f'(x) = \frac{(cx + d)(a) - (ax + b)(c)}{(cx + d)^2} = \frac{a(cx + d) - c(ax + b)}{(cx + d)^2}$$ Simplify numerator: $$a cx + a d - c a x - c b = a d - c b$$ So: $$f'(x) = \frac{a d - c b}{(cx + d)^2}$$ 9. Problem 116: $f(x) = \frac{1}{(1 + x^2)^2} = (1 + x^2)^{-2}$ Use chain rule: $$f'(x) = -2(1 + x^2)^{-3} \cdot 2x = -4x(1 + x^2)^{-3} = -\frac{4x}{(1 + x^2)^3}$$ 10. Problem 117: $f(x) = \frac{x}{1 + \sqrt{x}}$ Rewrite denominator: $$1 + x^{1/2}$$ Use quotient rule: $$f'(x) = \frac{(1 + x^{1/2})(1) - x \cdot \frac{1}{2} x^{-1/2}}{(1 + x^{1/2})^2} = \frac{1 + x^{1/2} - \frac{1}{2} x^{1/2}}{(1 + x^{1/2})^2} = \frac{1 + \frac{1}{2} x^{1/2}}{(1 + x^{1/2})^2}$$ 11. Problem 118: $f(x) = \sqrt{\frac{1 - x}{1 + x}} = \left(\frac{1 - x}{1 + x}\right)^{1/2}$ Use chain rule and quotient rule inside: Let $g(x) = \frac{1 - x}{1 + x}$ $$g'(x) = \frac{(1 + x)(-1) - (1 - x)(1)}{(1 + x)^2} = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2}$$ Then: $$f'(x) = \frac{1}{2} g(x)^{-1/2} \cdot g'(x) = \frac{1}{2} \left(\frac{1 - x}{1 + x}\right)^{-1/2} \cdot \left(-\frac{2}{(1 + x)^2}\right) = -\frac{1}{(1 + x)^2} \cdot \left(\frac{1 + x}{1 - x}\right)^{1/2}$$ Simplify: $$f'(x) = -\frac{\sqrt{1 + x}}{(1 + x)^2 \sqrt{1 - x}} = -\frac{1}{\sqrt{(1 - x)(1 + x)} (1 + x)} = -\frac{1}{(1 + x) \sqrt{1 - x^2}}$$ 12. Problem 119: $f(x) = \sqrt[3]{x + \sqrt{x}} = (x + x^{1/2})^{1/3}$ Use chain rule: $$f'(x) = \frac{1}{3} (x + x^{1/2})^{-2/3} \cdot \left(1 + \frac{1}{2} x^{-1/2}\right) = \frac{1 + \frac{1}{2} x^{-1/2}}{3 (x + \sqrt{x})^{2/3}}$$ 13. Problem 120: $\varphi(t) = \frac{t}{1 + \sqrt{t}}$ Use quotient rule: $$\varphi'(t) = \frac{(1 + t^{1/2})(1) - t \cdot \frac{1}{2} t^{-1/2}}{(1 + t^{1/2})^2} = \frac{1 + t^{1/2} - \frac{1}{2} t^{1/2}}{(1 + t^{1/2})^2} = \frac{1 + \frac{1}{2} t^{1/2}}{(1 + t^{1/2})^2}$$ 14. Problem 121: $g(s) = \sqrt{\frac{1 - s}{1 + s}}$ Same as problem 118, replace $x$ by $s$: $$g'(s) = -\frac{1}{(1 + s) \sqrt{1 - s^2}}$$ 15. Problem 122: $h(\rho) = \sqrt[3]{\rho + \sqrt{\rho}} = (\rho + \rho^{1/2})^{1/3}$ Use chain rule: $$h'(\rho) = \frac{1}{3} (\rho + \rho^{1/2})^{-2/3} \cdot \left(1 + \frac{1}{2} \rho^{-1/2}\right) = \frac{1 + \frac{1}{2} \rho^{-1/2}}{3 (\rho + \sqrt{\rho})^{2/3}}$$ 16. Group Problem 123: No graph provided, so no solution here. All problems solved step-by-step with simplifications.