Let $f\left( {x + y} \right) = f\left( x \right)f\left( y \right)$ and $f\left( x \right) = 1 + xg\left( x \right)G\left( x \right)$, where $\mathop {\lim }\limits_{x \to 0} g\left( x \right) = a$ and $\mathop {\lim }\limits_{x \to 0} G\left( x \right) = b$. Then $f'\left( x \right)$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

1 of 1

Analyzing +4 | -1

Let

f\left( {x + y} \right) = f\left( x \right)f\left( y \right)

and

f\left( x \right) = 1 + xg\left( x \right)G\left( x \right)

, where

\lim\limits_{x \to 0} g\left( x \right) = a

and

\lim\limits_{x \to 0} G\left( x \right) = b

. Then

f'\left( x \right)

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derivative

f'(x)

is equal to

ab f(x)

, which is not listed in the options.