Question

Given \[ \frac{d}{dx} \left( \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \right) = f(x) \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right] \] Then find \( f(5) \).

• A. \( \frac{72}{81e^5} \)
• B. \( \frac{7}{81e^5} \)
• C. \( \frac{8}{81e^5} \)
• D. \( e^5 \)

Hint:

Chain and Product Rule with Parameterization}
If \( \frac{d}{dx} f(x) = f(x) \cdot g(x) \), then original \( f(x) \) is isolated via algebra
Plug in value after full simplification
Watch out for squaring and radicals with specific values

Answer 1

The Correct Option is C

Observe the structure: Let: \[ f(x) = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \Rightarrow \text{Given that derivative } f'(x) = f(x) \cdot \left[\cdots\right] \] At \( x = 5 \), \[ f(5) = \frac{(6)^2 \cdot \sqrt{4}}{(9)^3 e^5} = \frac{36 \cdot 2}{729e^5} = \frac{72}{729e^5} = \frac{8}{81e^5} \]