6.1 Bepaal $f'(x)$ met gebruik van EERSTE BEGINSELS indien $f(x) = \frac{1}{2}^x$ - NSC Technical Mathematics - Question 6 - 2020 - Paper 1

For the area $A = \pi r^2$, the derivative with respect to $r$ is:

$\frac{dA}{dr} = 2\pi r$

This gives the rate of change of the area with respect to the radius.

To differentiate $D_{x}\left[\left(x - \sqrt{x}\right)^{2}\right]$, we apply the chain rule:

Let $u = x - \sqrt{x}$. Then:

$\frac{du}{dx} = 1 - \frac{1}{2\sqrt{x}}$

Thus,

Using the chain rule:

$D_{x}[u^2] = 2u \cdot \frac{du}{dx} = 2\left(x - \sqrt{x}\right)\left(1 - \frac{1}{2\sqrt{x}}\right)$

Thus, we find:

$D_{x}\left[\left(x - \sqrt{x}\right)^{2}\right] = 2\left(x - \sqrt{x}\right)\left(1 - \frac{1}{2\sqrt{x}}\right)$

Given the tangent line's equation $3x - y + 2 = 0$ and that it is tangent to the curve $g(x) = ax^2 - x$, we set:

y = $\frac{d}{dx}[g(x)]$ at the point $(-1, -1)$.

Given $g'(x) = 2ax - 1$, substituting $x = -1$:

$g'(-1) = 2a(-1) - 1 = -2a - 1$

Setting this equal to the slope from the line $3$:

$-2a - 1 = 3 \Rightarrow -2a = 4 \Rightarrow a = -2$

Thus, the value of $a$ is $-2$.