Bepaal $f'(x)$ met gebruik van EERSTE BEGINSELS as $f(x) = 5 + x$ (5) Bepaal: 6.2.1 $\frac{dy}{dx}$ as $y = x(x + 9)$ (3) 6.2.2 $D_x[\sqrt{x + \pi p^3}]$ (3) 6.2.3 $f'(x)$ as $f(x) = \frac{1 - x}{x^2}$ (4) Gegee $g(x) = -4x^2$ Bepaal: g(2) (1) 6.3.2 Die vergelyking van die raaklyn aan $g$ by 'n punt waar $x = 2$ (4) - NSC Technical Mathematics - Question 6 - 2022 - Paper 1

To find the derivative, we need to apply the product rule:

$y = x(x + 9)$

Applying the product rule:

$\frac{dy}{dx} = 1 \cdot (x + 9) + x \cdot 1 = x + 9 + x = 2x + 9$

Using differentiation rules for composite functions, we apply the chain rule:

Let $u = x + \pi p^3$, then:

$D_x[\sqrt{u}] = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$

Where:

$\frac{du}{dx} = 1$

Thus,

$D_x[\sqrt{x + \pi p^3}] = \frac{1}{2\sqrt{x + \pi p^3}}$

To calculate the derivative, we utilize the quotient rule:

$f'(x) = \frac{(x^2 \cdot (-1)) - (1 - x)(2x)}{(x^2)^2}$

Expanding gives:

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

To find $g(2)$, we substitute $x = 2$ into the function:

$g(2) = -4(2^2) = -4(4) = -16$

First, we find the derivative of $g(x)$:

$g'(x) = -8x$

At $x = 2$:

$g'(2) = -8(2) = -16$

Using the point-slope form of the equation of a line:

$y - g(2) = g'(2)(x - 2)$

We know $g(2) = -16$ and $g'(2) = -16$:

$y - (-16) = -16(x - 2)$

Thus, rearranging gives:

ightarrow y = -16x + 16$$