▶️ Answer/Explanation
Solution
First derivative: \( \frac{dy}{dx} = -6e^{-2x} \).
Second derivative: \( \frac{d^2 y}{dx^2} = 12e^{-2x} \).
Third derivative: \( \frac{d^3 y}{dx^3} = -24e^{-2x} \).
✅ Answer: A)
▶️ Answer/Explanation
Solution
We are given the graph of \( f'(x) \), the first derivative of \( f \).
The second derivative, \( f”(x) \), represents the slope of the graph of \( f'(x) \).
So to determine where \( f”(x) \) is least, we look for where the graph of \( f'(x) \) has the most negative slope (i.e., steepest descent).
Option A: At A, the graph has a horizontal tangent — the slope is 0. Therefore, \( f”(x) = 0 \).
Option B: At B, the graph of \( f'(x) \) is steepest downward — indicating the most negative slope. So \( f”(x) \) is smallest here.
Option C: At C, the slope is negative but less steep than at B, so \( f”(x) \) is negative but not the least.
Option D: At D, the slope is positive — the graph is increasing. So \( f”(x) > 0 \).
✅ Answer: B
▶️ Answer/Explanation
Solution
Differentiate: \( \frac{d^2 y}{dx^2} = \frac{2}{5} (x y^2 + 4y) \left( y^2 + (2x y + 4) \frac{dy}{dx} \right) \).
At \( x = -1 \), \( y = 5 \): \( \frac{dy}{dx} = 5 \), \( x y^2 + 4y = -5 \), \( y^2 + (2x y + 4) \frac{dy}{dx} = -5 \).
So, \( \frac{d^2 y}{dx^2} = \frac{2}{5} (-5) \cdot (-5) = 10 \).
✅ Answer: D)
▶️ Answer/Explanation
Solution
Derivatives: \( y’ = \cos x \), \( y” = -\sin x \), \( y”’ = -\cos x \), \( y^{(4)} = \sin x = y \).
The cycle repeats every 4 derivatives, so \( n = 4 \) is the smallest.
✅ Answer: B)