We are asked to solve four differentiation problems: a) Differentiate $y = 4x^2 - 2x$ from first principles. b) Differentiate $f(x) = 6 + \frac{1}{x^3}$ and find the gradient of the function when $x=2$. c) Differentiate $2x^3 \cos(3x)$. d) Differentiate $\frac{2x}{x^2+1}$.

1. Problem Description

We are asked to solve four differentiation problems:

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

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

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

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

2. Solution Steps

a) Differentiation from first principles:

The definition of the derivative from first principles is:

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

Given $y = f(x) = 4x^2 - 2x$, then $f(x+h) = 4(x+h)^2 - 2(x+h) = 4(x^2 + 2xh + h^2) - 2x - 2h = 4x^2 + 8xh + 4h^2 - 2x - 2h$.

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

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

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

b) Differentiation of $f(x) = 6 + \frac{1}{x^3}$ and gradient at $x=2$:

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

Then differentiate:

$f'(x) = 0 + (-3)x^{-4} = -3x^{-4} = -\frac{3}{x^4}$

Now, find the gradient at $x=2$:

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

c) Differentiation of $2x^3 \cos(3x)$:

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

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

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

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

d) Differentiation of $\frac{2x}{x^2+1}$:

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

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

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

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

3. Final Answer

a) $8x - 2$

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

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

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