Determine $f'(x)$ using FIRST PRINCIPLES if $f(x)=7x-2$ - NSC Technical Mathematics - Question 6 - 2018 - Paper 1

$\frac{d}{dx}(x^2) = 2x$.

Using the power rule:

$D_x (x^{\frac{3}{2}} - x^{\frac{1}{2}}) = \frac{3}{2} x^{\frac{3}{2}-1} - \frac{1}{2} x^{\frac{1}{2}-1}$

This simplifies to:

$= \frac{3}{2} x^{\frac{1}{2}} - \frac{1}{2} x^{-\frac{1}{2}} = \frac{3}{2\sqrt{x}} - \frac{1}{2} \frac{1}{\sqrt{x}} = \frac{2}{2\sqrt{x}} = \frac{2}{\sqrt{x}}$.

To differentiate $y = \frac{x^3 + 2}{x^2}$, we apply the quotient rule:

$\frac{dy}{dx} = \frac{(x^2)(3x^2) - (x^3 + 2)(2x)}{(x^2)^2}$

This gives:

$= \frac{3x^4 - (2x^4 + 4x)}{x^4} = \frac{x^4 - 4x}{x^4} = 1 - \frac{4}{x^3}$.

To find $k$, substitute $x=2$ in $p(x)$:

$$ p(2) = (2)^3 + 1 = 8 + 1 = k $,

Thus, $k = 9$.

Using the power rule for differentiation:

$$ p'(x) = \frac{d}{dx} (x^3 + 1) = 3x^2 $.

The derivative at point $A(2; 9)$ is:

$p'(2) = 3(2)^2 = 12.$

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$:

$y - 9 = 12(x - 2)$

This can be rearranged to:

$y = 12x - 24 + 9 = 12x - 15$,

Thus, the equation of the tangent line at point $A$ is $y = 12x - 15$.