6.1 Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = 2x + 3$ - NSC Technical Mathematics - Question 6 - 2021 - Paper 1

Using standard differentiation rules:

$\frac{dy}{dx} = -5x^4 + 12x^3$

We differentiate each term:

1. For $\frac{3}{x^4}$: $\frac{d}{dx}\left(3x^{-4}\right) = -12x^{-5}$

2. For $-\frac{x}{\sqrt{x}}$: $= -\frac{d}{dx}\left(x^{\frac{1}{2}}\right) = -\left(\frac{1}{2}x^{-\frac{1}{2}}\right) \Rightarrow -\frac{1}{2\sqrt{x}}$

Combining gives:

$f'(x) = -12x^{-5} - \frac{1}{2\sqrt{x}}$

To differentiate this quotient, we apply the quotient rule:

Let $u = x^2 + x - 6$ and $v = x + 3$.

The quotient rule states:

$\frac{D(u/v)}{dx} = \frac{v \cdot D(u) - u \cdot D(v)}{v^2}$

Calculating:

1. $D(u) = 2x + 1$
2. $D(v) = 1$

Thus,

$D_x \left[ \frac{x^2 + x - 6}{x + 3} \right] = \frac{(x + 3)(2x + 1) - (x^2 + x - 6)(1)}{(x + 3)^2}$

This simplifies to: $= 1$

The formula for average gradient between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m_{ave} = \frac{y_2 - y_1}{x_2 - x_1}$

Calculating y-values:

• For $x_1 = 0$: $y_1 = h(0) = 2$
• For $x_2 = 2$: $y_2 = h(2) = -2(2^2) + 2 = -6$

Thus,

$m_{ave} = \frac{-6 - 2}{2 - 0} = -4$

To find the gradient of the tangent, we first calculate $g'(-3)$.

Since $g(x) = 1 - x^2$, we differentiate:

$g'(x) = -2x$

Substituting $x = -3$ gives: $g'(-3) = -2(-3) = 6$

The equation of a tangent line can be expressed in point-slope form:

$y - y_1 = m(x - x_1)$

Where $(x_1, y_1) = (-3, g(-3))$. First calculate $g(-3)$:

$g(-3) = 1 - (-3)^2 = -8$

Thus, $y_1 = -8$, and $m = 6$. The equation becomes:

$y + 8 = 6(x + 3)$

Simplifying: $y = 6x + 18 - 8$ $y = 6x + 10$