1. Find the derivatives of each function using differentiation rules. (i) $f(x) = \sin (2x + 1)$ Using chain rule, $$f'(x) = \cos (2x + 1) \cdot \frac{d}{dx} (2x+1) = 2 \cos (2x+1)$$ (ii) $f(x) = 3 \cos \left(3x + \frac{\pi}{6}\right)$ $$f'(x) = 3 \cdot (-\sin (3x + \frac{\pi}{6})) \cdot 3 = -9 \sin \left(3x + \frac{\pi}{6}\right)$$ (iii) $f(x) = 2 \tan \left( \frac{x}{5} \right)$ $$f'(x) = 2 \sec^2 \left( \frac{x}{5} \right) \cdot \frac{1}{5} = \frac{2}{5} \sec^2 \left( \frac{x}{5} \right)$$ (iv) $f(x) = 4e^{5x^4+4}$ $$f'(x) = 4e^{5x^4+4} \cdot 20x^3 = 80x^3 e^{5x^4+4}$$ (v) $f(x) = \ln(2x + 3)$ $$f'(x) = \frac{1}{2x + 3} \cdot 2 = \frac{2}{2x + 3}$$ (vi) $f(x) = \ln(\sin x)$ $$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$$ (vii) $f(x) = \cos^2 x = (\cos x)^2$ Using chain rule, $$f'(x) = 2 \cos x (-\sin x) = -2 \sin x \cos x$$ (viii) $f(x) = \sin^3 (3x)$ $$f'(x) = 3 \sin^2 (3x) \cdot \cos (3x) \cdot 3 = 9 \sin^2 (3x) \cos (3x)$$ (ix) $f(x) = \tan^2 x$ $$f'(x) = 2 \tan x \cdot \sec^2 x$$ (x) $f(x) = e^{2 \sin (2x)}$ $$f'(x) = e^{2 \sin (2x)} \cdot 2 \cos (2x) \cdot 2 = 4 \cos (2x) e^{2 \sin (2x)}$$ (xi) $f(x) = e^{1 - 2 \tan (3x)}$ $$f'(x) = e^{1 - 2 \tan (3x)} \cdot (-2) \sec^2(3x) \cdot 3 = -6 \sec^2(3x) e^{1 - 2 \tan (3x)}$$ (xii) $f(x) = e^{2 \sin x + 5 \cos x}$ $$f'(x) = e^{2 \sin x + 5 \cos x} \cdot (2 \cos x - 5 \sin x)$$ 2. Find derivatives using product and quotient rules where applicable. (i) $f(x) = x \sin x$ $$f'(x) = \sin x + x \cos x$$ (ii) $f(x) = x^2 e^{3x}$ $$f'(x) = 2x e^{3x} + x^2 \cdot 3 e^{3x} = e^{3x}(2x + 3x^2)$$ (iii) $f(x) = x^3 \ln x$ $$f'(x) = 3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2 \ln x + x^2$$ (iv) $f(x) = e^{2x} \tan x$ $$f'(x) = 2 e^{2x} \tan x + e^{2x} \sec^2 x = e^{2x}(2 \tan x + \sec^2 x)$$ (v) $f(x) = \frac{x^2}{5x - 2}$ Using quotient rule, $$f'(x) = \frac{2x(5x - 2) - x^2 \cdot 5}{(5x - 2)^2} = \frac{2x(5x - 2) - 5x^2}{(5x - 2)^2} = \frac{10x^2 - 4x - 5x^2}{(5x - 2)^2} = \frac{5x^2 - 4x}{(5x - 2)^2}$$ (vi) $f(x) = \frac{\cos x}{x}$ $$f'(x) = \frac{-\sin x \cdot x - \cos x \cdot 1}{x^2} = \frac{-x \sin x - \cos x}{x^2}$$ (vii) $f(x) = \sec x$ $$f'(x) = \sec x \tan x$$ (viii) $f(x) = \csc x$ $$f'(x) = -\csc x \cot x$$ (ix) $f(x) = \cot x$ $$f'(x) = -\csc^2 x$$ (x) $f(x) = e^{-x} \sin x$ $$f'(x) = -e^{-x} \sin x + e^{-x} \cos x = e^{-x} (-\sin x + \cos x)$$ (xi) $f(x) = \frac{\sin^2 x}{x}$ Using quotient rule, $$f'(x) = \frac{2 \sin x \cos x \cdot x - \sin^2 x \cdot 1}{x^2} = \frac{2x \sin x \cos x - \sin^2 x}{x^2}$$ (xii) $f(x) = x \ln (\cos x)$ $$f'(x) = \ln (\cos x) + x \cdot \frac{-\sin x}{\cos x} = \ln (\cos x) - x \tan x$$ 3. Find smallest positive $x$ where $f'(x) = 0$ (stationary points): (i) $f(x) = 2x + 3 \cos x$ $$f'(x) = 2 - 3 \sin x = 0 \implies \sin x = \frac{2}{3}$$ Smallest positive solution, $$x = \arcsin \left( \frac{2}{3} \right)$$ (ii) $f(x) = x - 2 \sin x$ $$f'(x) = 1 - 2 \cos x = 0 \implies \cos x = \frac{1}{2}$$ Smallest positive value, $$x = \frac{\pi}{3}$$ (iii) $f(x) = 3 \sin x + 5 \cos x$ $$f'(x) = 3 \cos x - 5 \sin x = 0 \implies 3 \cos x = 5 \sin x \implies \tan x = \frac{3}{5}$$ Smallest positive solution, $$x = \arctan \left( \frac{3}{5} \right)$$ 4. Find $x$ for stationary points $f'(x) = 0$: (i) $f(x) = 3x - e^{2x}$ $$f'(x) = 3 - 2 e^{2x} = 0 \implies e^{2x} = \frac{3}{2} \implies 2x = \ln \frac{3}{2} \implies x = \frac{1}{2} \ln \frac{3}{2}$$ (ii) $f(x) = x^2 + 3 \ln x$ $$f'(x)= 2x + \frac{3}{x} = 0 \implies 2x^2 + 3 = 0$$ No real positive solution (discard negative $x$), so no stationary values for real $x > 0$. (iii) $f(x) = \ln(1/x) + 8x = - \ln x + 8x$ $$f'(x) = - \frac{1}{x} + 8 = 0 \implies 8 = \frac{1}{x} \implies x = \frac{1}{8}$$ 5. Find gradient and tangent/normal equations: (i) $y = \sin x + \cos x$, at $x = \frac{\pi}{2}$ Gradient, $$y' = \cos x - \sin x$$ At $x=\frac{\pi}{2}$, $$y' = \cos \frac{\pi}{2} - \sin \frac{\pi}{2} = 0 - 1 = -1$$ Point coordinate, $$y= \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$$ Tangent line, $$y - 1 = -1 (x - \frac{\pi}{2}) \implies y = -x + \frac{\pi}{2} + 1$$ Normal line slope = $\frac{1}{1} = 1$ Normal, $$y - 1 = 1 (x - \frac{\pi}{2}) \implies y = x - \frac{\pi}{2} +1$$ (ii) $y = x + e^x$, $x = 0$ $$y' = 1 + e^x$$ At $x=0$, $$y' = 1 + 1 = 2$$ Point, $$y=0 +1=1$$ Tangent, $$y - 1 = 2 (x - 0) \implies y = 2x +1$$ Normal slope = $-\frac{1}{2}$, Normal, $$y -1 = -\frac{1}{2} (x-0) \implies y = - \frac{1}{2} x + 1$$ (iii) $y = 2 + x^2 + \ln x$, $x=1$ $$y' = 2x + \frac{1}{x} = 2(1) + 1 = 3$$ Point, $$y = 2 + 1 + 0 = 3$$ Tangent, $$y - 3 = 3(x - 1) \implies y = 3x \text{ (simplified)}$$ Normal slope = $-\frac{1}{3}$ Normal, $$y - 3 = -\frac{1}{3}(x - 1) \implies y = -\frac{1}{3} x + \frac{10}{3}$$ (iv) $y = x \sin x$ at $x=\frac{\pi}{2}$ $$y' = \sin x + x \cos x$$ At $x=\frac{\pi}{2}$, $$y' = 1 + \frac{\pi}{2} \cdot 0 = 1$$ Point, $$y = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$$ Tangent, $$y - \frac{\pi}{2} = 1 (x - \frac{\pi}{2}) \implies y = x$$ Normal slope = $-1$ Normal, $$y - \frac{\pi}{2} = -1 (x - \frac{\pi}{2}) \implies y = -x + \pi$$ 6. For $y = 4x - 2 \cos x$, find point where tangent slope = 5 $$y' = 4 + 2 \sin x = 5 \implies 2 \sin x = 1 \implies \sin x = \frac{1}{2}$$ Smallest positive $x = \frac{\pi}{6}$ Coordinate, $$y = 4 \cdot \frac{\pi}{6} - 2 \cos \frac{\pi}{6} = \frac{2 \pi}{3} - 2 \cdot \frac{\sqrt{3}}{2} = \frac{2 \pi}{3} - \sqrt{3}$$ 7. For $y = 3 \ln x + x$, normal parallel to line $2x + 3y =4$ Slope of line, $$- \frac{2}{3}$$ Gradient of curve, $$y' = \frac{3}{x} + 1$$ Slope of normal, $$- \frac{1}{y'} = - \frac{1}{\frac{3}{x} + 1} = - \frac{1}{\frac{3 + x}{x}} = - \frac{x}{3 + x}$$ Set equal to slope of line, $$- \frac{x}{3 + x} = - \frac{2}{3} \implies \frac{x}{3 + x} = \frac{2}{3}$$ Cross multiply, $$3x = 2(3 + x) = 6 + 2x \implies x = 6$$ Coordinates, $$y = 3 \ln 6 + 6$$ 8. Stationary point for $y = (x + 3) e^{-x}$ $$y' = e^{-x} (1 + x) - (x + 3) e^{-x} = e^{-x} (1 + x - x - 3) = e^{-x} (-2) = -2 e^{-x}$$ Set to zero, $$-2 e^{-x} = 0$$ No solution; derivative never zero. Check carefully: Rewrite derivative correctly: $$y' = (1) e^{-x} + (x + 3)(-e^{-x}) = e^{-x} - (x + 3) e^{-x} = e^{-x}(1 - x - 3) = e^{-x} (-x - 2)$$ Set to zero, $$e^{-x} (-x - 2) = 0 \implies -x - 2 = 0 \implies x = -2$$ Second derivative test: $$y'' = \frac{d}{dx} \left(e^{-x} (-x - 2) \right) = - e^{-x} (-x - 2) + e^{-x} (-1) = e^{-x} ( -1 + x + 2) = e^{-x} (x + 1)$$ At $x = -2$, $$y''(-2) = e^{2} (-1) < 0$$ So maximum at $x = -2$. 9. Stationary point for $$y = \frac{e^{2x}}{\cos x}$$ Derivative, $$y' = \frac{2 e^{2x} \cos x + e^{2x} \sin x}{\cos^2 x} = \frac{e^{2x}(2 \cos x + \sin x)}{\cos^2 x}$$ Set numerator zero, $$2 \cos x + \sin x = 0 \implies 2 \cos x = - \sin x \implies \tan x = -2$$ $x$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$, $$x = \arctan(-2)$$ Coordinates, $$y = \frac{e^{2 \arctan(-2)}}{\cos \arctan(-2)}$$ 10. Stationary point of $$y= \frac{e^x}{x -1}$$ Derivative, $$y' = \frac{e^x (x-1) - e^x}{(x-1)^2} = \frac{e^x (x-1 -1)}{(x-1)^2} = \frac{e^x (x-2)}{(x-1)^2}$$ Set numerator zero, $$x - 2 = 0 \implies x = 2$$ Second derivative test shows minimum: Final stationary point coordinates, $$\left(2, \frac{e^2}{2-1} \right) = (2, e^2)$$