The problem consists of several calculus questions: a) Differentiate $f(x) = x^2$ from first principles and find the gradient at $x=2$. b) Find the derivatives of: (i) $f(x) = 6x^2 - 9x + 4$ and (ii) $y = \sqrt{x} + 8\sqrt[3]{x} - 2\sqrt[4]{x}$. c) Use the quotient rule to find the derivative of $g(x) = \frac{6x^2}{2-x}$. d) Differentiate $f(x) = 2e^x - 8^x$. e) Integrate $\int (2 + \frac{5}{7}x - 6x^2) dx$. f) Evaluate the definite integral $\int_1^4 (3x-2) dx$. g) Evaluate $\int \frac{3x+11}{x^2 - x - 6} dx$ using partial fractions.

1. Problem Description

The problem consists of several calculus questions:

a) Differentiate $f(x) = x^2$ from first principles and find the gradient at $x=2$.

b) Find the derivatives of: (i) $f(x) = 6x^2 - 9x + 4$ and (ii) $y = \sqrt{x} + 8\sqrt[3]{x} - 2\sqrt[4]{x}$.

c) Use the quotient rule to find the derivative of $g(x) = \frac{6x^2}{2-x}$.

d) Differentiate $f(x) = 2e^x - 8^x$.

e) Integrate $\int (2 + \frac{5}{7}x - 6x^2) dx$.

f) Evaluate the definite integral $\int_1^4 (3x-2) dx$.

g) Evaluate $\int \frac{3x+11}{x^2 - x - 6} dx$ using partial fractions.

2. Solution Steps

a) Differentiation from first principles:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

For $f(x) = x^2$, $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$.

$f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x$.

The derivative is $f'(x) = 2x$. At $x=2$, the gradient is $f'(2) = 2(2) = 4$.

b) (i) $f(x) = 6x^2 - 9x + 4$

$f'(x) = 12x - 9$.

(ii) $y = \sqrt{x} + 8\sqrt[3]{x} - 2\sqrt[4]{x} = x^{1/2} + 8x^{1/3} - 2x^{1/4}$

$y' = \frac{1}{2}x^{-1/2} + \frac{8}{3}x^{-2/3} - \frac{2}{4}x^{-3/4} = \frac{1}{2\sqrt{x}} + \frac{8}{3\sqrt[3]{x^2}} - \frac{1}{2\sqrt[4]{x^3}}$.

c) Quotient rule: If $g(x) = \frac{u(x)}{v(x)}$, then $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

$g(x) = \frac{6x^2}{2-x}$. Let $u(x) = 6x^2$ and $v(x) = 2-x$. Then $u'(x) = 12x$ and $v'(x) = -1$.

$g'(x) = \frac{(12x)(2-x) - (6x^2)(-1)}{(2-x)^2} = \frac{24x - 12x^2 + 6x^2}{(2-x)^2} = \frac{24x - 6x^2}{(2-x)^2} = \frac{6x(4-x)}{(2-x)^2}$.

d) $f(x) = 2e^x - 8^x$

$f'(x) = 2e^x - 8^x \ln(8)$.

e) $\int (2 + \frac{5}{7}x - 6x^2) dx = 2x + \frac{5}{7} \cdot \frac{x^2}{2} - 6 \cdot \frac{x^3}{3} + C = 2x + \frac{5}{14}x^2 - 2x^3 + C$.

f) $\int_1^4 (3x-2) dx = [\frac{3}{2}x^2 - 2x]_1^4 = (\frac{3}{2}(4^2) - 2(4)) - (\frac{3}{2}(1^2) - 2(1)) = (24 - 8) - (\frac{3}{2} - 2) = 16 - (-\frac{1}{2}) = 16 + \frac{1}{2} = \frac{33}{2}$.

g) $\int \frac{3x+11}{x^2 - x - 6} dx = \int \frac{3x+11}{(x-3)(x+2)} dx$.

Using partial fractions, $\frac{3x+11}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$.

$3x+11 = A(x+2) + B(x-3)$.

When $x=3$, $3(3)+11 = A(3+2) + B(3-3) \Rightarrow 20 = 5A \Rightarrow A=4$.

When $x=-2$, $3(-2)+11 = A(-2+2) + B(-2-3) \Rightarrow 5 = -5B \Rightarrow B=-1$.

$\int \frac{3x+11}{(x-3)(x+2)} dx = \int (\frac{4}{x-3} - \frac{1}{x+2}) dx = 4\ln|x-3| - \ln|x+2| + C$.

3. Final Answer

a) $f'(x) = 2x$, Gradient at $x=2$ is

4. b) (i) $f'(x) = 12x - 9$ (ii) $y' = \frac{1}{2\sqrt{x}} + \frac{8}{3\sqrt[3]{x^2}} - \frac{1}{2\sqrt[4]{x^3}}$

c) $g'(x) = \frac{6x(4-x)}{(2-x)^2}$

d) $f'(x) = 2e^x - 8^x \ln(8)$

e) $\int (2 + \frac{5}{7}x - 6x^2) dx = 2x + \frac{5}{14}x^2 - 2x^3 + C$

f) $\int_1^4 (3x-2) dx = \frac{33}{2}$

g) $\int \frac{3x+11}{x^2 - x - 6} dx = 4\ln|x-3| - \ln|x+2| + C$