Given: $$ f(x) = 11 + 7x $$ Determine $f' (x)$ using FIRST PRINCIPLES: --- 6.2 Determine: 6.2.1 $$ \frac{dy}{dx} $$ if $$ y = x^* $$ 6.2.2 $$ f' (x) $$ if $$ f(x)=\sqrt{x} $$ 6.2.3 $$ D_{x} \left[ \frac{x^2 - 16}{4 - x} \right] $$ --- 6.3 Determine the average gradient of the function defined by $$ g(x) = \frac{9}{x} $$ between $$ x = -3 $$ and $$ x = -1 $$ --- Given: $$ f(x) = mx + m - 4 $$ 6.4.1 Determine $$ f' (x) $$ in terms of $$ m $$ 6.4.2 Hence, calculate $$ f' (2) $$ in terms of $$ m $$ 6.4.3 Determine the numerical value of $$ m $$ if the gradient of the tangent to $$ f $$ at $$ x = 2 $$ is equal to 39. - NSC Technical Mathematics - Question 6 - 2024 - Paper 1

$\frac{dy}{dx} = 8x^7.$

Using the power rule, we can write: $f(x) = x^{1/2}$
Thus, $f'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}.$

Using the quotient rule: $D_{x}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}.$
With $u = x^2 - 16$ and $v = 4 - x$, we find: $u' = 2x, \quad v' = -1$
Thus, $D_{x}\left[\frac{x^2 - 16}{4 - x}\right] = \frac{(2x)(4 - x) - (x^2 - 16)(-1)}{(4 - x)^2}.$
This simplifies to: $D_{x}\left[\frac{x^2 - 16}{4 - x}\right] = \frac{(2x(4 - x) + x^2 - 16)}{(4 - x)^2} = -1.$

To find the average gradient, we calculate: $g(-3) = -3 \text{ and } g(-1) = -9.$
Thus, $\text{Av. Gradient} = \frac{g(x_2) - g(x_1)}{x_2 - x_1} = \frac{-9 - (-3)}{-1 - (-3)} = \frac{-6}{2} = -3.$

Given $f(x) = mx + m - 4$, we differentiate: $f' (x) = m.$

$f' (2) = 3m.$

Setting the derivative equal to 39: $3m = 39$
Solving for $m$ gives: $m = 13.$