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

To differentiate $y = x(x + 9)$, we first expand the expression:

$y = x^2 + 9x$

Now, differentiate:

$\frac{dy}{dx} = 2x + 9$

To differentiate $\sqrt{x + \pi p^3}$, we apply the chain rule:

$D_x[\sqrt{x + \pi p^3}] = \frac{1}{2\sqrt{x + \pi p^3}} \cdot \frac{d}{dx}(x + \pi p^3) = \frac{1}{2\sqrt{x + \pi p^3}}$

To differentiate $f(x)$, we apply the quotient rule:

Let $u = 1 - x^9$ and $v = x^2$:

$f'(x) = \frac{u'v - uv'}{v^2}$

Calculating $u' = -9x^8$ and $v' = 2x$:

$f'(x) = \frac{(-9x^8)(x^2) - (1 - x^9)(2x)}{x^4}$

Simplifying:

$f'(x) = \frac{-9x^{10} - 2x + 2x^{10}}{x^4} = \frac{-7x^{10} - 2x}{x^4} = -7x^6 - \frac{2}{x^3}$

To evaluate $g(2)$ where $g(x) = -4x^2$:

$g(2) = -4(2^2) = -4(4) = -16$

First, we find the derivative of $g(x)$:

$g'(x) = -8x$

Evaluating at $x = 2$:

$g'(2) = -8(2) = -16$

Now, we use the point-slope form to write the equation of the tangent line:

Let the point be $(2, g(2)) = (2, -16)$:

$y - (-16) = -16(x - 2)$

This simplifies to:

$y = -16x + 16 - 16 = -16x + 0$

Thus, the equation of the tangent line is:

$y = -16x$