The problem consists of five parts, each requiring differentiation. a) Differentiate $y = 4x^2 - 2x$ from first principles. b) Differentiate $f(x) = (6 + \frac{1}{x^3})$ and find the gradient at $x = 2$. c) Differentiate $2x^3 \cos(3x)$. d) Differentiate $\frac{2x}{x^2 + 1}$. e) Differentiate $(2x^3 - 5x)^5$.

1. Problem Description

The problem consists of five parts, each requiring differentiation.

a) Differentiate $y = 4x^2 - 2x$ from first principles.

b) Differentiate $f(x) = (6 + \frac{1}{x^3})$ and find the gradient at $x = 2$.

c) Differentiate $2x^3 \cos(3x)$.

d) Differentiate $\frac{2x}{x^2 + 1}$.

e) Differentiate $(2x^3 - 5x)^5$.

2. Solution Steps

a) Differentiate $y = 4x^2 - 2x$ from first principles.

The definition of derivative from first principles is:

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

Here, $f(x) = 4x^2 - 2x$. So,

$f(x+h) = 4(x+h)^2 - 2(x+h) = 4(x^2 + 2xh + h^2) - 2x - 2h = 4x^2 + 8xh + 4h^2 - 2x - 2h$

$f(x+h) - f(x) = (4x^2 + 8xh + 4h^2 - 2x - 2h) - (4x^2 - 2x) = 8xh + 4h^2 - 2h$

$\frac{f(x+h) - f(x)}{h} = \frac{8xh + 4h^2 - 2h}{h} = 8x + 4h - 2$

$f'(x) = \lim_{h \to 0} (8x + 4h - 2) = 8x - 2$

b) Differentiate $f(x) = (6 + \frac{1}{x^3})$ and find the gradient at $x = 2$.

First rewrite the function as $f(x) = 6 + x^{-3}$.

Then differentiate:

$f'(x) = -3x^{-4} = -\frac{3}{x^4}$

At $x = 2$, the gradient is:

$f'(2) = -\frac{3}{2^4} = -\frac{3}{16}$

c) Differentiate $2x^3 \cos(3x)$.

Use the product rule: $(uv)' = u'v + uv'$

Let $u = 2x^3$ and $v = \cos(3x)$

$u' = 6x^2$ and $v' = -3\sin(3x)$

$(2x^3 \cos(3x))' = (6x^2) \cos(3x) + (2x^3)(-3\sin(3x)) = 6x^2 \cos(3x) - 6x^3 \sin(3x)$

d) Differentiate $\frac{2x}{x^2 + 1}$.

Use the quotient rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$

Let $u = 2x$ and $v = x^2 + 1$

$u' = 2$ and $v' = 2x$

$(\frac{2x}{x^2 + 1})' = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}$

e) Differentiate $(2x^3 - 5x)^5$.

Use the chain rule: $(f(g(x)))' = f'(g(x))g'(x)$

Let $f(u) = u^5$ and $g(x) = 2x^3 - 5x$

$f'(u) = 5u^4$ and $g'(x) = 6x^2 - 5$

$((2x^3 - 5x)^5)' = 5(2x^3 - 5x)^4 (6x^2 - 5)$

3. Final Answer

a) $8x - 2$

b) $-\frac{3}{16}$

c) $6x^2 \cos(3x) - 6x^3 \sin(3x)$

d) $\frac{2 - 2x^2}{(x^2 + 1)^2}$

e) $5(2x^3 - 5x)^4 (6x^2 - 5)$