Grade 12 Mathematics "Differential Calculus"

here are the answers with detailed workings for the Grade 12 Mathematics "Differential Calculus" questions:

1. Differentiate the function:
   \[
   f(x) = 3x^4 - 5x^2 + 2x - 7
   \]

   **Solution:**
   \[
   f'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(-7)
   \]
   \[
   f'(x) = 12x^3 - 10x + 2
   \]

2. Find the derivative of the function:
   \[
   g(x) = \sin(3x) + \cos(2x)
   \]

   **Solution:**
   \[
   g'(x) = \frac{d}{dx}(\sin(3x)) + \frac{d}{dx}(\cos(2x))
   \]
   \[
   g'(x) = 3\cos(3x) - 2\sin(2x)
   \]

3. If \( h(x) = e^{2x} \cdot \ln(x) \), determine \( h'(x) \) using the product rule.

   **Solution:**
   \[
   h'(x) = \frac{d}{dx}(e^{2x}) \cdot \ln(x) + e^{2x} \cdot \frac{d}{dx}(\ln(x))
   \]
   \[
   h'(x) = 2e^{2x} \cdot \ln(x) + e^{2x} \cdot \frac{1}{x}
   \]
   \[
   h'(x) = 2e^{2x} \ln(x) + \frac{e^{2x}}{x}
   \]

4. Given \( f(x) = \frac{2x^3 - 4x^2 + x}{x^2} \), simplify the function and then find \( f'(x) \).

   **Solution:**
   \[
   f(x) = \frac{2x^3}{x^2} - \frac{4x^2}{x^2} + \frac{x}{x^2}
   \]
   \[
   f(x) = 2x - 4 + \frac{1}{x}
   \]
   \[
   f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}(4) + \frac{d}{dx}\left(\frac{1}{x}\right)
   \]
   \[
   f'(x) = 2 - \frac{1}{x^2}
   \]

5. Determine the critical points of the function:
   \[
   f(x) = x^3 - 6x^2 + 9x + 1
   \]

   **Solution:**
   \[
   f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 1)
   \]
   \[
   f'(x) = 3x^2 - 12x + 9
   \]
   Set \( f'(x) = 0 \) to find critical points:
   \[
   3x^2 - 12x + 9 = 0
   \]
   \[
   x^2 - 4x + 3 = 0
   \]
   \[
   (x - 1)(x - 3) = 0
   \]
   \[
   x = 1 \text{ or } x = 3
   \]

6. For the function \( f(x) = \ln(x^2 + 1) \), find the second derivative, \( f''(x) \).

   **Solution:**
   \[
   f'(x) = \frac{d}{dx}(\ln(x^2 + 1))
   \]
   \[
   f'(x) = \frac{2x}{x^2 + 1}
   \]
   Now, find \( f''(x) \):
   \[
   f''(x) = \frac{d}{dx}\left(\frac{2x}{x^2 + 1}\right)
   \]
   \[
   f''(x) = \frac{(2)(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2}
   \]
   \[
   f''(x) = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2}
   \]
   \[
   f''(x) = \frac{-2x^2 + 2}{(x^2 + 1)^2}
   \]
   \[
   f''(x) = \frac{2(1 - x^2)}{(x^2 + 1)^2}
   \]

7. Solve the differential equation:
   \[
   \frac{dy}{dx} = 5x^4 - 3x^2 + 2
   \]

   **Solution:**
   Integrate both sides with respect to \( x \):
   \[
   y = \int (5x^4 - 3x^2 + 2) \, dx
   \]
   \[
   y = \frac{5}{5}x^5 - \frac{3}{3}x^3 + 2x + C
   \]
   \[
   y = x^5 - x^3 + 2x + C
   \]

8. Find the equation of the tangent line to the curve \( y = x^2 - 4x + 4 \) at the point \( (2, 0) \).

   **Solution:**
   First, find the derivative:
   \[
   y' = \frac{d}{dx}(x^2 - 4x + 4)
   \]
   \[
   y' = 2x - 4
   \]
   At \( x = 2 \):
   \[
   y' = 2(2) - 4 = 0
   \]
   The slope of the tangent line is 0, so the equation of the tangent line is horizontal at \( y = 0 \):
   \[
   y = 0
   \]

9. Determine the local maxima and minima for the function:
   \[
   f(x) = 2x^3 - 9x^2 + 12x - 4
   \]

   **Solution:**
   First, find the derivative:
   \[
   f'(x) = 6x^2 - 18x + 12
   \]
   Set \( f'(x) = 0 \) to find critical points:
   \[
   6x^2 - 18x + 12 = 0
   \]
   \[
   x^2 - 3x + 2 = 0
   \]
   \[
   (x - 1)(x - 2) = 0
   \]
   \[
   x = 1 \text{ or } x = 2
   \]
   Find the second derivative to determine concavity:
   \[
   f''(x) = 12x - 18
   \]
   At \( x = 1 \):
   \[
   f''(1) = 12(1) - 18 = -6 \quad (\text{concave down, local maximum})
   \]
   At \( x = 2 \):
   \[
   f''(2) = 12(2) - 18 = 6 \quad (\text{concave up, local minimum})
   \]

10. Evaluate the following limit using L'Hôpital's rule:
    \[
    \lim_{{x \to 0}} \frac{\sin(x)}{x}
    \]

    **Solution:**
    Using L'Hôpital's rule:
    \[
    \lim_{{x \to 0}} \frac{\sin(x)}{x} = \lim_{{x \to 0}} \frac{\cos(x)}{1} = \cos(0) = 1
    \]

Feel free to ask if you need any further explanations or have any more questions!