1. **Problem statement:** Differentiate the following functions: (i) $f(x) = 4x^5 - 5x^4$ (ii) $f(x) = (x^6 + 1)^5 (4x + 7)^3$ (iii) $f(x) = \left(7x + \sqrt{x^2 + 3}\right)^6$ (iv) $f(x) = \frac{(x - 1)^3}{x (x + 3)^5}$ --- 2. **Formulas and rules used:** - Power rule: $\frac{d}{dx} x^n = n x^{n-1}$ - Product rule: $\frac{d}{dx} [u v] = u' v + u v'$ - Chain rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ - Quotient rule: $\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u' v - u v'}{v^2}$ - Derivative of $\sqrt{x} = x^{1/2}$ is $\frac{1}{2 \sqrt{x}}$ --- 3. **Step-by-step solutions:** (i) Differentiate $f(x) = 4x^5 - 5x^4$ $$f'(x) = 4 \cdot 5 x^{5-1} - 5 \cdot 4 x^{4-1} = 20 x^4 - 20 x^3$$ (ii) Differentiate $f(x) = (x^6 + 1)^5 (4x + 7)^3$ Let $u = (x^6 + 1)^5$, $v = (4x + 7)^3$ Calculate $u'$ using chain rule: $$u' = 5 (x^6 + 1)^4 \cdot 6 x^5 = 30 x^5 (x^6 + 1)^4$$ Calculate $v'$ using chain rule: $$v' = 3 (4x + 7)^2 \cdot 4 = 12 (4x + 7)^2$$ Apply product rule: $$f'(x) = u' v + u v' = 30 x^5 (x^6 + 1)^4 (4x + 7)^3 + (x^6 + 1)^5 12 (4x + 7)^2$$ (iii) Differentiate $f(x) = \left(7x + \sqrt{x^2 + 3}\right)^6$ Let $g(x) = 7x + \sqrt{x^2 + 3}$ First find $g'(x)$: $$\frac{d}{dx} 7x = 7$$ $$\frac{d}{dx} \sqrt{x^2 + 3} = \frac{1}{2 \sqrt{x^2 + 3}} \cdot 2x = \frac{x}{\sqrt{x^2 + 3}}$$ So, $$g'(x) = 7 + \frac{x}{\sqrt{x^2 + 3}}$$ Now apply chain rule: $$f'(x) = 6 \left(7x + \sqrt{x^2 + 3}\right)^5 \cdot \left(7 + \frac{x}{\sqrt{x^2 + 3}}\right)$$ (iv) Differentiate $f(x) = \frac{(x - 1)^3}{x (x + 3)^5}$ Let numerator $u = (x - 1)^3$, denominator $v = x (x + 3)^5$ Calculate $u'$: $$u' = 3 (x - 1)^2$$ Calculate $v'$ using product rule: Let $a = x$, $b = (x + 3)^5$ $$a' = 1$$ $$b' = 5 (x + 3)^4$$ So, $$v' = a' b + a b' = (1) (x + 3)^5 + x \cdot 5 (x + 3)^4 = (x + 3)^4 \left( (x + 3) + 5x \right) = (x + 3)^4 (6x + 3)$$ Apply quotient rule: $$f'(x) = \frac{u' v - u v'}{v^2} = \frac{3 (x - 1)^2 \cdot x (x + 3)^5 - (x - 1)^3 (x + 3)^4 (6x + 3)}{\left[x (x + 3)^5\right]^2}$$ Simplify numerator by factoring $(x - 1)^2 (x + 3)^4$: $$= \frac{(x - 1)^2 (x + 3)^4 \left[3 x (x + 3) - (x - 1)(6x + 3)\right]}{x^2 (x + 3)^{10}}$$ Simplify inside bracket: $$3 x (x + 3) = 3x^2 + 9x$$ $$(x - 1)(6x + 3) = 6x^2 + 3x - 6x - 3 = 6x^2 - 3x - 3$$ So bracket becomes: $$3x^2 + 9x - (6x^2 - 3x - 3) = 3x^2 + 9x - 6x^2 + 3x + 3 = -3x^2 + 12x + 3$$ Final derivative: $$f'(x) = \frac{(x - 1)^2 (x + 3)^4 (-3x^2 + 12x + 3)}{x^2 (x + 3)^{10}} = \frac{(x - 1)^2 (-3x^2 + 12x + 3)}{x^2 (x + 3)^6}$$ --- **Final answers:** (i) $f'(x) = 20 x^4 - 20 x^3$ (ii) $f'(x) = 30 x^5 (x^6 + 1)^4 (4x + 7)^3 + 12 (x^6 + 1)^5 (4x + 7)^2$ (iii) $f'(x) = 6 \left(7x + \sqrt{x^2 + 3}\right)^5 \left(7 + \frac{x}{\sqrt{x^2 + 3}}\right)$ (iv) $f'(x) = \frac{(x - 1)^2 (-3x^2 + 12x + 3)}{x^2 (x + 3)^6}$