Answer is: option4

-2

Solution:

Since \( \ln e^{2x} = 2x \), we simplify: \[ f(x) = \frac{2x}{x - 1} \] The derivative of an inverse function is given by: \[ g'(y) = \frac{1}{f'(g(y))} \]

We need to compute \( f'(x) \) and evaluate it at \( x = g(3) \).

Using the quotient rule: \[ f(x) = \frac{2x}{x - 1} \]

Let \( u = 2x \) and \( v = x - 1 \), then:

\[ f'(x) = \frac{(2)(x - 1) - (2x)(1)}{(x - 1)^2} \] \[ = \frac{2x - 2 - 2x}{(x - 1)^2} \] \[ = \frac{-2}{(x - 1)^2} \] Find \( x \) such that \( f(x) = 3 \): \[ \frac{2x}{x - 1} = 3 \]

Cross multiplying:

\[ 2x = 3(x - 1) \] \[ 2x = 3x - 3 \] \[ x = 3 \]

Thus, \( g(3) = 3 \).

Now, compute \( f'(3) \): \[ f'(3) = \frac{-2}{(3 - 1)^2} = \frac{-2}{4} = -\frac{1}{2} \] Using the inverse function derivative formula: \[ g'(3) = \frac{1}{f'(3)} = \frac{1}{-1/2} = -2 \] Final Answer: \[ \boxed{-2} \]