1. **Find the derivative of** $f(x) = \sqrt[3]{x^2} + \frac{3x}{x^4} - \frac{1}{2 \sqrt[4]{x^5}} - 6$. Do not simplify. Step 1: Rewrite the function using exponents: $$f(x) = x^{\frac{2}{3}} + 3x \cdot x^{-4} - \frac{1}{2} \, x^{-\frac{5}{4}} - 6$$ Step 2: Simplify terms inside the function: $$f(x) = x^{\frac{2}{3}} + 3x^{-3} - \frac{1}{2} x^{-\frac{5}{4}} - 6$$ Step 3: Differentiate term-by-term using the power rule $\frac{d}{dx} x^n = n x^{n-1}$ and constants rule: $$f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} + 3 \cdot (-3) x^{-3 - 1} - \frac{1}{2} \cdot \left(-\frac{5}{4}\right) x^{-\frac{5}{4} - 1} - 0$$ Step 4: Write the derivative explicitly: $$f'(x) = \frac{2}{3} x^{-\frac{1}{3}} - 9 x^{-4} + \frac{5}{8} x^{-\frac{9}{4}}$$ 2. **Find the derivative of** $f(x) = \frac{\sin(\tan x)}{\sqrt[3]{\cos x}}$. Do not simplify. Step 1: Set numerator $u = \sin(\tan x)$ and denominator $v = (\cos x)^{1/3}$. Step 2: Differentiate numerator using the chain rule: $$u' = \cos(\tan x) \cdot \sec^2 x$$ Step 3: Differentiate denominator: $$v = (\cos x)^{1/3} \Rightarrow v' = \frac{1}{3} (\cos x)^{-\frac{2}{3}} (-\sin x) = - \frac{\sin x}{3 (\cos x)^{2/3}}$$ Step 4: Use the quotient rule: $$f'(x) = \frac{u' v - u v'}{v^2} = \frac{\cos(\tan x) \sec^2 x (\cos x)^{1/3} - \sin(\tan x) \left(- \frac{\sin x}{3 (\cos x)^{2/3}}\right)}{(\cos x)^{2/3}}$$ 3. **Find the derivative of** $f(x) = \frac{(2x - 1)^2 (3x^2 + 2x - 1)^3}{(4x^3 - 5x^2 + 11)^4}$. Do not simplify. Step 1: Let $u = (2x - 1)^2 (3x^2 + 2x - 1)^3$ and $v = (4x^3 - 5x^2 + 11)^4$. Step 2: Differentiate $u$ using product rule: $$u = p \cdot q \text{ where } p = (2x -1)^2, q = (3x^2 + 2x -1)^3$$ $$p' = 2 (2x - 1) \cdot 2 = 4(2x - 1)$$ $$q' = 3 (3x^2 + 2x -1)^2 (6x + 2)$$ $$u' = p' q + p q' = 4(2x - 1)(3x^2 + 2x - 1)^3 + (2x - 1)^2 \cdot 3(3x^2 + 2x - 1)^2 (6x + 2)$$ Step 3: Differentiate $v$ using chain rule: $$v' = 4 (4x^3 - 5x^2 + 11)^3 (12x^2 - 10x)$$ Step 4: Use quotient rule for $f(x) = \frac{u}{v}$: $$f'(x) = \frac{u' v - u v'}{v^2}$$ 4. **Find the equation of the tangent and normal lines to** $f(x) = \frac{2 \sqrt{x - 1}}{x^2 + 1}$ at $x = 2$. Step 1: Write $f(x)$: $$f(x) = \frac{2 (x-1)^{1/2}}{x^2 + 1}$$ Step 2: Find $f(2)$: $$f(2) = \frac{2 (2-1)^{1/2}}{2^2 + 1} = \frac{2 \cdot 1}{4 + 1} = \frac{2}{5}$$ Step 3: Compute $f'(x)$ using quotient rule: Let $u = 2 (x-1)^{1/2}$ and $v = x^2 + 1$. $$u' = 2 \cdot \frac{1}{2} (x-1)^{-1/2} = (x-1)^{-1/2}$$ $$v' = 2x$$ $$f'(x) = \frac{u' v - u v'}{v^2} = \frac{(x-1)^{-1/2} (x^2 + 1) - 2 (x-1)^{1/2} (2x)}{(x^2 + 1)^2}$$ Step 4: Evaluate $f'(2)$: $$f'(2) = \frac{(1)^{-1/2} (4 + 1) - 2 (1)^{1/2} (4)}{(5)^2} = \frac{5 - 8}{25} = -\frac{3}{25}$$ Step 5: Equation of the tangent line at $(2, \frac{2}{5})$: $$y - \frac{2}{5} = -\frac{3}{25} (x - 2)$$ Step 6: Slope of normal line is negative reciprocal: $$m_{normal} = \frac{25}{3}$$ Equation of normal line: $$y - \frac{2}{5} = \frac{25}{3} (x - 2)$$ 5. **Find point(s) on graph of** $f(x)=x^5+4x-3$ where tangent line passes through $(0,1)$. Step 1: Derivative: $$f'(x) = 5x^4 + 4$$ Step 2: Equation of tangent line at $x = a$: $$y = f'(a)(x - a) + f(a)$$ It passes through $(0,1)$ so substitute $x=0$, $y=1$: $$1 = f'(a) (0 - a) + f(a) = -a f'(a) + f(a)$$ Step 3: Write out $f(a)$ and $f'(a)$: $$1 = -a (5a^4 + 4) + (a^5 + 4a - 3) = -5a^5 - 4a + a^5 + 4a - 3 = -4a^5 - 3$$ Step 4: Solve for $a$: $$1 = -4a^5 - 3 \Rightarrow 4 = -4a^5 \Rightarrow a^5 = -1$$ $$a = -1$$ Step 5: The point is: $$f(-1) = (-1)^5 + 4(-1) - 3 = -1 - 4 - 3 = -8$$ So the point is $(-1, -8)$. 6. **Find** $a, b \in \mathbb{R}$ **such that** $$f(x) = \begin{cases} a x^2 + b x + 1 & x \leq 1 \\ a x + 2b & x > 1 \end{cases}$$ is differentiable at $x=1$. Step 1: For continuity at $x=1$: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$ Evaluate left hand limit: $$a(1)^2 + b(1) + 1 = a + b + 1$$ Evaluate right hand limit: $$a(1) + 2b = a + 2b$$ Set equal for continuity: $$a + b + 1 = a + 2b \Rightarrow b + 1 = 2b \Rightarrow b = 1$$ Step 2: For differentiability, derivatives from left and right must be equal at $x=1$. Left derivative: $$f'_-(1) = 2a(1) + b = 2a + b$$ Right derivative: $$f'_+(1) = a$$ Set equal: $$2a + b = a \Rightarrow a + b = 0$$ Using $b=1$: $$a + 1 = 0 \Rightarrow a = -1$$