Determine $f' (x)$ using FIRST PRINCIPLES if $f (x) = 5 - 8x$ - NSC Technical Mathematics - Question 6 - 2022 - Paper 1

To find $f' (x)$, we differentiate both terms:

• The derivative of $3x^3$ is $9x^2$
• The derivative of rac{\pi}{x} is $-\frac{\pi}{x^2}$, using the power rule or quotient rule.

Combining these results gives: $f' (x) = 9x^2 - \frac{\pi}{x^2}$

For this, we apply the product rule: Let:

• $u = x^2$ and $v = (4x - 2x^2)$.

The derivatives are:

• $\frac{du}{dx} = 2x$
• $\frac{dv}{dx} = 4 - 4x$.

Using the product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$ This results in: $\frac{dy}{dx} = x^2(4 - 4x) + (4x - 2x^2)(2x)$ Simplifying gives: $\frac{dy}{dx} = 4x^2 - 4x^3 + 8x^2 - 4x^3 = 12x^2 - 8x^3$

Here, we need to differentiate the expression with respect to $x$. Let's denote: $y = \sqrt{4x^4 - \frac{2}{5x}} + 8r^*x$ The derivative can be found via the chain and sum rules. For the first part: $\frac{d}{dx} \left(\sqrt{4x^4 - \frac{2}{5x}}\right) = \frac{1}{2\sqrt{4x^4 - \frac{2}{5x}}} \cdot (16x^3 + \frac{2}{5x^2})$ For the second part, differentiate $8r^*x$: $\frac{d}{dx}(8r^*x) = 8r^*$ Then combine both results to form: D_x \left{ y \right} = \frac{1}{2\sqrt{4x^4 - \frac{2}{5x}}} \cdot (16x^3 + \frac{2}{5x^2}) + 8r^*

First, differentiate $g (x)$ to find $g' (x)$: $g' (x) = 12x + 3$ Setting this equal to -21 to find $p$ gives: $12p + 3 = -21$ Solving for $p$: $12p = -24 \Rightarrow p = -2$

Using the point-slope form for the tangent line, we have:

1. The slope at $p = -2$ is: $g' (-2) = 12(-2) + 3 = -24 + 3 = -21$
2. The point on the curve at $x = -2$ is: $g (-2) = 6(-2)^2 + 3(-2) = 24 - 6 = 18$
3. The equation of the tangent line is given by: $y - 18 = -21(x + 2)$ Rearranging yields: $y = -21x - 42 + 18$ Thus, the equation is: $y = -21x - 24$