6.1 Determine f'(x) using FIRST PRINCIPLES if f(x) = 5 - \frac{1}{2} x 6.2 Determine the following: 6.2.1 f'(x) if f(x) = a^3 - 0.5x^{-1} 6.2.2 D_x \left[ x \left( \sqrt{x+2} \right) \right] 6.3 Given: xy + 2x^3y = 7x^6 6.3.1 Make y the subject of the equation - NSC Technical Mathematics - Question 6 - 2019 - Paper 1

Differentiating the function, we find: $f'(x) = 0 - (-0.5)x^{-2} = 0.5x^{-2}$ So, $f'(x) = \frac{0.5}{x^2}$.

To differentiate the function, we use the product rule: If $u = x$ and $v = \sqrt{x+2}$, then: $D_x [uv] = u'v + uv'$ Calculating $u' = 1$ and $v' = \frac{1}{2}(x+2)^{-1/2}$Thus:$D_x \left[ x \left( \sqrt{x+2} \right) \right] = 1 \cdot \sqrt{x+2} + x \cdot \left(\frac{1}{2}(x+2)^{-1/2}\right)$Simplifying:$D_x \left[ x \left( \sqrt{x+2} \right) \right] = \sqrt{x+2} + \frac{x}{2\sqrt{x+2}} = \frac{2(x+2) + x}{2\sqrt{x+2}} = \frac{3x + 4}{2\sqrt{x+2}}$$

Starting with the equation: $xy + 2x^3y = 7x^6$ Factoring out $y$ provides: $y(x + 2x^3) = 7x^6$ Isolating $y$ gives us: $y = \frac{7x^6}{x + 2x^3}$

Using the quotient rule for differentiation: $\frac{dy}{dx} = \frac{(7x^6)'}{x + 2x^3} - \frac{7x^6 (x + 2x^3)'}{(x + 2x^3)^2}$ Calculating: $(7x^6)' = 42x^5$ $(x + 2x^3)' = 1 + 6x^2$ Combining gives: $\frac{dy}{dx} = \frac{42x^5 (x + 2x^3) - 7x^6 (1 + 6x^2)}{(x + 2x^3)^2}$

Substituting $x = 300$ into the profit function: $P(300) = 0.8(300)^2 - 200(300) = 0.8(90000) - 60000 = 72000 - 60000 = 12000\text{ rands}$

Setting the profit equation to zero: $0 = 0.8x^2 - 200x$ Factoring out $x$ gives: $x(0.8x - 200) = 0$ This yields two solutions: $x = 0 \text{ or } x = \frac{200}{0.8} = 250$ Thus, 250 light bulbs will yield zero profit.

To find the derivative of the profit function: $P'(x) = (0.8x^2 - 200x)' = 1.6x - 200$ Substituting $x = 200$ results in: $P'(200) = 1.6(200) - 200 = 320 - 200 = 120\text{ bulbs per day}$