1. Find the derivative of $y = \frac{\sin x}{\cos x}$. Use the quotient rule: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$$ where $u = \sin x$ and $v = \cos x$. Calculate derivatives: $u' = \cos x$, $v' = -\sin x$. Apply quotient rule: $$y' = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$ 2. Find the derivative of $y = \sin(3x) \cos(2x)$. Use product rule: $$\frac{d}{dx}(fg) = f'g + fg'$$ with $f = \sin(3x)$ and $g = \cos(2x)$. Calculate derivatives: $f' = 3\cos(3x)$, $g' = -2\sin(2x)$. Apply product rule: $$y' = 3\cos(3x) \cdot \cos(2x) + \sin(3x) \cdot (-2\sin(2x)) = 3\cos(3x)\cos(2x) - 2\sin(3x)\sin(2x)$$ 3. Find the derivative of $y = \frac{\sin^2 x + \cos^2 x}{\tan x}$. Recall identity: $\sin^2 x + \cos^2 x = 1$, so $y = \frac{1}{\tan x} = \cot x$. Derivative of $\cot x$ is: $$y' = -\csc^2 x$$ 4. Find the derivative of $y = 2 \tan(3x) + \cos(5x)$. Use chain rule: $$\frac{d}{dx} \tan(3x) = 3 \sec^2(3x)$$ $$\frac{d}{dx} \cos(5x) = -5 \sin(5x)$$ Therefore: $$y' = 2 \cdot 3 \sec^2(3x) - 5 \sin(5x) = 6 \sec^2(3x) - 5 \sin(5x)$$ 5. Find the derivative of $y = x^2 \sin x - 3x \cos x + 5 \sin x$. Use product rule for $x^2 \sin x$ and $-3x \cos x$. Derivatives: $$\frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x$$ $$\frac{d}{dx}(-3x \cos x) = -3 \cos x + 3x \sin x$$ $$\frac{d}{dx}(5 \sin x) = 5 \cos x$$ Sum all: $$y' = 2x \sin x + x^2 \cos x - 3 \cos x + 3x \sin x + 5 \cos x = (2x \sin x + 3x \sin x) + (x^2 \cos x - 3 \cos x + 5 \cos x) = 5x \sin x + (x^2 + 2) \cos x$$ 6. Find the derivative of $y = \ln(\cos(2x))$. Use chain rule: $$y' = \frac{1}{\cos(2x)} \cdot (-\sin(2x)) \cdot 2 = -2 \tan(2x)$$ 7. Find the third derivative of $y = -4x^2 e^x + 5x e^x - 10 e^x$. First derivative: $$y' = -4(2x e^x + x^2 e^x) + 5(e^x + x e^x) - 10 e^x = -8x e^x - 4x^2 e^x + 5 e^x + 5x e^x - 10 e^x$$ Simplify: $$y' = (-8x + 5x - 4x^2) e^x + (5 - 10) e^x = (-4x^2 - 3x - 5) e^x$$ Second derivative: $$y'' = \frac{d}{dx}((-4x^2 - 3x - 5) e^x) = (-8x - 3) e^x + (-4x^2 - 3x - 5) e^x = (-4x^2 - 11x - 8) e^x$$ Third derivative: $$y''' = \frac{d}{dx}((-4x^2 - 11x - 8) e^x) = (-8x - 11) e^x + (-4x^2 - 11x - 8) e^x = (-4x^2 - 19x - 19) e^x$$ --- Part 2: 1. Given parametric equations $x(t) = t^2$, $y(t) = 2t + 1$, find $\frac{dy}{dx}$. Use chain rule: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2t} = \frac{1}{t}$$ 2. Find the point of inflection for $y = x^4 - 2x^2$. Find second derivative: $$y' = 4x^3 - 4x$$ $$y'' = 12x^2 - 4$$ Set $y'' = 0$: $$12x^2 - 4 = 0 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm \frac{1}{\sqrt{3}}$$ Points of inflection at $x = \pm \frac{1}{\sqrt{3}}$. 3. Find equations of tangent and normal lines to $y = \sqrt{x + 1}$ at point $(8, 3)$. Derivative: $$y' = \frac{1}{2 \sqrt{x + 1}}$$ At $x=8$: $$y' = \frac{1}{2 \sqrt{9}} = \frac{1}{6}$$ Tangent line slope $m_t = \frac{1}{6}$. Equation tangent line: $$y - 3 = \frac{1}{6}(x - 8)$$ Normal line slope $m_n = -6$ (negative reciprocal). Equation normal line: $$y - 3 = -6(x - 8)$$ 4. Find equations of tangent and normal lines to $y = \sqrt[4]{x}$ at point $(16, 2)$. Rewrite $y = x^{1/4}$. Derivative: $$y' = \frac{1}{4} x^{-3/4}$$ At $x=16$: $$y' = \frac{1}{4} (16)^{-3/4} = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}$$ Tangent line slope $m_t = \frac{1}{32}$. Equation tangent line: $$y - 2 = \frac{1}{32}(x - 16)$$ Normal line slope $m_n = -32$. Equation normal line: $$y - 2 = -32(x - 16)$$ 5. Find equation of normal line for implicit function $2xy^2 + y^2 = 10$ at point $(1, 5)$. Differentiate implicitly: $$\frac{d}{dx}(2xy^2) + \frac{d}{dx}(y^2) = 0$$ Use product rule on $2xy^2$: $$2(y^2 + x \cdot 2y \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0$$ Simplify: $$2y^2 + 4xy \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$$ Group $\frac{dy}{dx}$ terms: $$(4xy + 2y) \frac{dy}{dx} = -2y^2$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{-2y^2}{4xy + 2y} = \frac{-2y^2}{2y(2x + 1)} = \frac{-y}{2x + 1}$$ At $(1,5)$: $$\frac{dy}{dx} = \frac{-5}{2(1) + 1} = \frac{-5}{3}$$ Normal line slope is negative reciprocal: $$m_n = \frac{3}{5}$$ Equation of normal line: $$y - 5 = \frac{3}{5}(x - 1)$$