JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

f(x)=\left(\frac{a+x}{b+x}\right)^{a+b+2 x}

, then

f^{\prime}(0)

is equal to

View answer and explanation

Correct option: B

\left(2 \log \frac{a}{b}+\frac{b^2-a^2}{a b}\right)\left(\frac{a}{b}\right)^{a+b}

Option B (

\left(2 \log \frac{a}{b}+\frac{b^2-a^2}{a b}\right)\left(\frac{a}{b}\right)^{a+b}

) is correct because it is the result of applying logarithmic differentiation and evaluating at x=0.

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