6.1 Determine $f' (x)$ using FIRST PRINCIPLES if $f (x) = \frac{1}{2} x$ - NSC Technical Mathematics - Question 6 - 2020 - Paper 1

To compute the derivative of area $A$ with respect to radius $r$, we use:

$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r.$

To find the derivative $D_x [ (x - \sqrt{x})^2 ]$, we will use the chain rule:

Let $u = x - \sqrt{x}$, then:

$D_x [u^2] = 2u\cdot D_x[u]$ where:

$D_x[u] = \frac{d}{dx}(x - x^{1/2}) = 1 - \frac{1}{2}x^{-1/2} = 1 - \frac{1}{2\sqrt{x}}.$

Thus,

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

To find the point of contact, we substitute the point $(-1, -1)$ into the function:

$g(-1) = a(-1)^2 - (-1) = a + 1.$

Setting this equal to $-1$ (the $y$ coordinate of the point of contact):

$a + 1 = -1$

From which we solve:

$a = -2.$