Question

If $f(1) = 3,\; f'(1) = 2$, then $\dfrac{d}{dx}\left\{\log\left[f\left(e^x + 2x\right)\right]\right\}$ at $x = 0$ is ______.

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $2$
• D. $0$

Hint:

Always differentiate outside-in using the chain rule.

Answer 1

The Correct Option is C

Step 1: Chain Rule
Let $y = \log [ f(e^x + 2x) ]$.
$\frac{dy}{dx} = \frac{1}{f(e^x + 2x)} \cdot f'(e^x + 2x) \cdot (e^x + 2)$.
Step 2: Substitute $x=0$
At $x=0$, $e^0 + 2(0) = 1$.
Value $= \frac{1}{f(1)} \cdot f'(1) \cdot (e^0 + 2)$.
Step 3: Calculation
Value $= \frac{1}{3} \cdot 2 \cdot (1 + 2) = \frac{2}{3} \times 3 = 2$.
Final Answer: (C)