We need to solve five differentiation problems: 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

We need to solve five differentiation problems:

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:

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

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

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

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

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

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

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

Now, differentiate: $f'(x) = -3x^{-4} = -\frac{3}{x^4}$.

To find the gradient at $x = 2$, substitute $x = 2$ into the derivative:

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

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

Use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.

Here, $u = 2x^3$ and $v = \cos(3x)$.

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

$\frac{d}{dx}(2x^3\cos(3x)) = (6x^2)\cos(3x) + (2x^3)(-3\sin(3x)) = 6x^2\cos(3x) - 6x^3\sin(3x)$.

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

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

Here, $u = 2x$ and $v = x^2 + 1$.

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

$\frac{d}{dx}(\frac{2x}{x^2 + 1}) = \frac{(2)(x^2 + 1) - (2x)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}$.

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

Use the chain rule: $\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$.

Here, $f(u) = u^5$ and $g(x) = 2x^3 - 5x$.

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

$\frac{d}{dx}((2x^3 - 5x)^5) = 5(2x^3 - 5x)^4(6x^2 - 5)$.

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