Question

If \( f(x) = \left(\frac{3+x}{1+x}\right)^{2+3x} \), then \( f'(0) = \)

• A. \( 12 + \log 3 \)
• B. \( -12 + 3\log 3 \)
• C. \( -\frac{4}{3} + 3\log 3 \)
• D. \( -12 + 27\log 3 \)

Hint:

For functions of the type \( y = (f(x))^{g(x)} \), use logarithmic differentiation for easier handling.

Answer 1

The Correct Option is D

Step 1: Take logarithm of the function.
Let:
\[ y = \left(\frac{3+x}{1+x}\right)^{2+3x}. \]
Taking logarithm:
\[ \log y = (2+3x)\log\left(\frac{3+x}{1+x}\right). \]

Step 2: Differentiate implicitly.
\[ \frac{1}{y}\frac{dy}{dx} = (2+3x)\frac{d}{dx}\log\left(\frac{3+x}{1+x}\right) + 3\log\left(\frac{3+x}{1+x}\right). \]

Step 3: Differentiate the log term.
\[ \log\left(\frac{3+x}{1+x}\right) = \log(3+x) - \log(1+x). \]
\[ \frac{d}{dx} = \frac{1}{3+x} - \frac{1}{1+x}. \]

Step 4: Substitute into derivative expression.
\[ \frac{1}{y}\frac{dy}{dx} = (2+3x)\left(\frac{1}{3+x} - \frac{1}{1+x}\right) + 3\log\left(\frac{3+x}{1+x}\right). \]

Step 5: Evaluate at \( x = 0 \).
First compute \( y \) at \( x=0 \):
\[ y = \left(\frac{3}{1}\right)^2 = 9. \]
Now evaluate RHS:
\[ (2)\left(\frac{1}{3} - 1\right) + 3\log 3. \]
\[ = 2\left(-\frac{2}{3}\right) + 3\log 3. \]
\[ = -\frac{4}{3} + 3\log 3. \]

Step 6: Multiply by \( y \).
\[ \frac{dy}{dx} = y \left(-\frac{4}{3} + 3\log 3\right). \]
\[ = 9\left(-\frac{4}{3} + 3\log 3\right). \]
\[ = -12 + 27\log 3. \]

Step 7: Final conclusion.
Thus,
\[ f'(0) = -12 + 27\log 3. \]
Final Answer:
\[ \boxed{-12 + 27\log 3}. \]