The function `f(x)=e^x+x ,` being differentiable and one-to-one, has a differentiable inverse `f^(-1)(x)dot` The value of `d/(dx)(f^(-1))` at the point `f(log2)` is `1/(1n2)` (b) `1/3` (c) `1/4` (d) none of these

Options

A. `(1)/(In2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. none of these

Explanation

To find the value of \( \frac{d}{dx}(f^{-1}(x)) \) at the point \( f(\log 2) \), we can follow these steps:

### Step 1: Define the function and its inverse
Let \( f(x) = e^x + x \). Since \( f \) is a differentiable and one-to-one function, it has an inverse \( f^{-1}(x) \).

### Step 2: Use the relationship between the derivatives of a function and its inverse
By the inverse function theorem, we have:
\[
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