The problem asks us to find derivatives of different functions. Specifically, a) Differentiate $f(x) = x^2$ from first principles and find the gradient at $x=2$. b) Find the derivatives of $f(x) = 6x^2 - 9x + 4$ and $y = \sqrt{x} + 8\sqrt[3]{x} - 2\sqrt[4]{x}$. c) Use the quotient rule to find the derivative of $g(x) = \frac{6x^2}{2-x}$. d) Differentiate $f(x) = 2e^x - 8^x$.

1. Problem Description

The problem asks us to find derivatives of different functions. Specifically,

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

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

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

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

2. Solution Steps

a) Differentiating from first principles:

The definition of the derivative from first principles is:

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

For $f(x) = x^2$:

$f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$

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

So, $f'(x) = 2x$.

The gradient at $x=2$ is $f'(2) = 2(2) = 4$.

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

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

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

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

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

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

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

Here, $u(x) = 6x^2$, $u'(x) = 12x$

$v(x) = 2-x$, $v'(x) = -1$

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

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

The derivative of $e^x$ is $e^x$. The derivative of $a^x$ is $a^x \ln a$.

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

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