7.1 Given: f(x) = 2x^2 - x Determine f'(x) from first principles - NSC Mathematics - Question 7 - 2017 - Paper 1

To differentiate the product of two functions, we apply the product rule:

$D_x[uv] = u'v + uv'$

Let:

• u = (x + 1) → u' = 1
• v = (3x - 7) → v' = 3

Now applying the product rule: $D_x[(x + 1)(3x - 7)] = (1)(3x - 7) + (x + 1)(3)$

Which simplifies to: $D_x[(x + 1)(3x - 7)] = 3x - 7 + 3x + 3 = 6x - 4$

To find the derivative of the given function, we first express it clearly:

$y = \frac{\sqrt{x - 5}}{x} \cdot \frac{1}{2\pi}$

Using the quotient rule for differentiation:

$\frac{dy}{dx} = \frac{(u'v - uv')}{v^2}$

Let:

• u = √(x - 5) → u' = \frac{1}{2\sqrt{x - 5}}
• v = x → v' = 1

Now substituting into the quotient rule: $\frac{dy}{dx} = \frac{\left(\frac{1}{2\sqrt{x - 5}} \cdot x - \sqrt{x - 5} \cdot 1\right)}{x^2} \cdot \frac{1}{2\pi}$

This gives: $\frac{dy}{dx} = \frac{1}{2\pi}\cdot \frac{x - 2\sqrt{x - 5}}{2x^2\sqrt{x - 5}}$