Let `f(x)=int_(2)^(x)(dt)/(sqrt(1+t^(4)))` and `g` be the inverse of `f` then the value of `g^(')(0)` is

Options

A. `1`
B. `17`
C. `sqrt(17)`
D. `0`

Explanation

To find the value of \( g'(0) \) where \( g \) is the inverse of the function \( f(x) = \int_{2}^{x} \frac{dt}{\sqrt{1 + t^4}} \), we can use the relationship between the derivatives of inverse functions. Here's a step-by-step solution:

### Step 1: Understand the relationship between \( f \) and \( g \)

Since \( g \) is the inverse of \( f \), we have:
\[ g(f(x)) = x \]
Differentiating both sides with respect to \( x \) gives us:
\[ g'(f(x)) \cdot f'(x) = 1 \]

### Step 2: Find \( f'(x) \)

Using the Fundamental Theorem of Calculus, we can find \( f'(x) \):
\[ f'(x) = \frac{d}{dx} \left( \int_{2}^{x} \frac{dt}{\sqrt{1 + t^4}} \right) = \frac{1}{\sqrt{1 + x^4}} \]

### Step 3: Evaluate \( f(2) \)

Next, we need to find \( f(2) \):
\[ f(2) = \int_{2}^{2} \frac{dt}{\sqrt{1 + t^4}} = 0 \]

### Step 4: Use the relationship to find \( g'(0) \)

Since \( f(2) = 0 \), we can substitute \( x = 2 \) into the derivative relationship:
\[ g'(f(2)) \cdot f'(2) = 1 \]
This simplifies to:
\[ g'(0) \cdot f'(2) = 1 \]

### Step 5: Calculate \( f'(2) \)

Now we substitute \( x = 2 \) into \( f'(x) \):
\[ f'(2) = \frac{1}{\sqrt{1 + 2^4}} = \frac{1}{\sqrt{1 + 16}} = \frac{1}{\sqrt{17}} \]

### Step 6: Solve for \( g'(0) \)

Now we can solve for \( g'(0) \):
\[ g'(0) \cdot \frac{1}{\sqrt{17}} = 1 \]
Thus,
\[ g'(0) = \sqrt{17} \]

### Final Answer

The value of \( g'(0) \) is:
\[ \boxed{\sqrt{17}} \]