7.1 Bepaal $f'(x)$ vanuit eerste beginsels indien $f(x)=2x^2-1$ - NSC Mathematics - Question 7 - 2020 - Paper 1

To differentiate $\sqrt{x^2 + x^3}$, we can use the chain rule. Let $u = x^2 + x^3$, then:

1. Differentiate using chain rule:
   $\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$

2. Calculate $\frac{du}{dx}$:
   $\frac{du}{dx} = 2x + 3x^2$

3. Therefore,
   $\frac{d}{dx}(\sqrt{x^2+x^3}) = \frac{1}{2\sqrt{x^2+x^3}} \cdot (2x + 3x^2)$
   The final answer is
   $\frac{2x + 3x^2}{2\sqrt{x^2 + x^3}}$

To find the derivative of $f(x) = \frac{4x^2 - 9}{4x + 6}$, we can use the quotient rule:

1. Differentiate using the quotient rule:
   $f'(x) = \frac{(g(x) \cdot f'(x) - f(x) \cdot g'(x))}{(g(x))^2}$
   where $f(x) = 4x^2 - 9$ and $g(x) = 4x + 6$.

2. Compute $f'(x)$ and $g'(x)$:

   • $f'(x) = 8x$
   • $g'(x) = 4$

3. Applying the quotient rule gives: $f'(x) = \frac{(4x + 6)(8x) - (4x^2 - 9)(4)}{(4x + 6)^2}$

4. To find $f'(\frac{3}{2})$, substitute $x = \frac{3}{2}$: $f'\left(\frac{3}{2}\right) = \frac{(4 \cdot \frac{3}{2} + 6)(8 \cdot \frac{3}{2}) - (4\left(\frac{3}{2}\right)^2 - 9)(4)}{(4 \cdot \frac{3}{2} + 6)^2}$

5. Calculate to find the answer.