If f(x)=≤ft(3+x/1+x)²+3 x}, then f^{}(0)=

Correct Answer is : −12+27log⁡3-12+27 3−12+27log3

Correct Answer is : −12+27log⁡3-12+27 3−12+27log3

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COMEDK 2025 Morning Solved Paper

Question : 140 of 180

Marks: +1, -0

f(x)=\left(\frac{3+x}{1+x}\right)^{2+3 x}

f^{\prime}(0)=

Solution:

Concept:

Use logarithmic differentiation for functions of the form

f(x) = g(x)^{h(x)}

Explanation:

Let

g(x) = \frac{3+x}{1+x}

and

h(x) = 2+3x

, so

f(x) = g(x)^{h(x)}

Take natural log:

\ln f(x) = h(x) \ln g(x)

Differentiate:

\frac{f'(x)}{f(x)} = h'(x) \ln g(x) + h(x) \cdot \frac{g'(x)}{g(x)}

Thus

f'(x) = f(x) \left( h'(x) \ln g(x) + h(x) \cdot \frac{g'(x)}{g(x)} \right)

Evaluate at

x=0

g(0) = \frac{3+0}{1+0} = 3

h(0) = 2+0 = 2

f(0) = 3^2 = 9

g'(x) = \frac{(1+x) - (3+x)}{(1+x)^2} = \frac{-2}{(1+x)^2}

, so

g'(0) = -2

h'(x) = 3

, so

h'(0) = 3

Substitute into the formula:

f'(0) = 9 \left( 3 \cdot \ln 3 + 2 \cdot \frac{-2}{3} \right) = 9 \left( 3 \ln 3 - \frac{4}{3} \right) = 27 \ln 3 - 12

Therefore

f'(0) = -12 + 27 \log 3

(using

\log

for natural log).

Answer:

Option D:

-12 + 27 \log 3

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