Let g be a real valued function defined on the interval (-1, 1) such that `e^(-x) (g (x) - 2e^(x)) = underset(0)overset(x)intsqrt(y^(4) +1) dy` for all ` in (-1, 1)` and f be an another function such f(g(x)) = g(f(x)) = x. Then the value of f'(2) is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(1)/(5)`
D. `(1)/(3)`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the given equation
We start with the equation:
\[ e^{-x} (g(x) - 2e^{x}) = \int_0^x \sqrt{y^4 + 1} \, dy \]
This equation holds for all \( x \) in the interval \( (-1, 1) \).

### Step 2: Differentiate both sides
We differentiate both sides with respect to \( x \):
- Left-hand side:
\[ \frac{d}{dx}\left(e^{-x} (g(x) - 2e^{x})\right) = e^{-x} (g'(x) - 2e^{x}) + (g(x) - 2e^{x})(-e^{-x}) \]
This simplifies to:
\[ e^{-x} (g'(x) - 2e^{x}) - e^{-x} (g(x) - 2e^{x}) = e^{-x} (g'(x) - g(x)) \]

- Right-hand side:
Using the Fundamental Theorem of Calculus, we differentiate the integral:
\[ \frac{d}{dx}\left(\int_0^x \sqrt{y^4 + 1} \, dy\right) = \sqrt{x^4 + 1} \]

Thus, we have:
\[ e^{-x} (g'(x) - g(x)) = \sqrt{x^4 + 1} \]

### Step 3: Rearranging the equation
We can rearrange this to express \( g'(x) \):
\[ g'(x) - g(x) = e^{x} \sqrt{x^4 + 1} \]
So,
\[ g'(x) = g(x) + e^{x} \sqrt{x^4 + 1} \]

### Step 4: Finding \( g(0) \)
Next, we need to find \( g(0) \). We substitute \( x = 0 \) into the original equation:
\[ e^{0} (g(0) - 2e^{0}) = \int_0^0 \sqrt{y^4 + 1} \, dy \]
This simplifies to:
\[ g(0) - 2 = 0 \implies g(0) = 2 \]

### Step 5: Finding \( g'(0) \)
Now we substitute \( x = 0 \) into the differentiated equation:
\[ g'(0) = g(0) + e^{0} \sqrt{0^4 + 1} \]
Substituting \( g(0) = 2 \):
\[ g'(0) = 2 + 1 = 3 \]

### Step 6: Using the relationship between \( f \) and \( g \)
We know that:
\[ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x \]
Differentiating \( f(g(x)) = x \) gives:
\[ f'(g(x)) g'(x) = 1 \]
At \( x = 0 \):
\[ f'(g(0)) g'(0) = 1 \implies f'(2) \cdot 3 = 1 \]
Thus,
\[ f'(2) = \frac{1}{3} \]

### Final Answer
The value of \( f'(2) \) is:
\[ \boxed{\frac{1}{3}} \]