Let `f(x)=x+ sin x` . Suppose g denotes the inverse function of f. The value of `g'(pi/4+1/sqrt2)` has the value equal to

Options

A. `sqrt(2)-1`
B. `(sqrt(2)+1)/(sqrt(2))`
C. `2-sqrt(2)`
D. `sqrt(2)+1`

Answer

To find the value of \( g'(\frac{\pi}{4} + \frac{1}{\sqrt{2}}) \) where \( g \) is the inverse function of \( f(x) = x + \sin x \), we will use the formula for the derivative of the inverse function.

### Step-by-Step Solution:

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 \( x \):
\[ g'(f(x)) \cdot f'(x) = 1 \]
This implies:
\[ g'(f(x)) = \frac{1}{f'(x)} \]

2. **Find \( f'(x) \)**:
The function \( f(x) = x + \sin x \) can be differentiated:
\[ f'(x) = 1 + \cos x \]

3. **Set up the equation for \( g' \)**:
We need to find \( g'(\frac{\pi}{4} + \frac{1}{\sqrt{2}}) \). Let:
\[ y = \frac{\pi}{4} + \frac{1}{\sqrt{2}} \]
Then, we need to find \( x \) such that:
\[ f(x) = y \]
This means:
\[ x + \sin x = \frac{\pi}{4} + \frac{1}{\sqrt{2}} \]

4. **Find \( x \)**:
We can try \( x = \frac{\pi}{4} \):
\[ f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \sin\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{1}{\sqrt{2}} \]
This confirms that \( x = \frac{\pi}{4} \) is the correct value.

5. **Calculate \( g'(\frac{\pi}{4} + \frac{1}{\sqrt{2}}) \)**:
Now we can substitute \( x = \frac{\pi}{4} \) into the formula for \( g' \):
\[ g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right) = \frac{1}{f'\left(\frac{\pi}{4}\right)} \]

6. **Evaluate \( f'\left(\frac{\pi}{4}\right) \)**:
\[ f'\left(\frac{\pi}{4}\right) = 1 + \cos\left(\frac{\pi}{4}\right) = 1 + \frac{1}{\sqrt{2}} \]

7. **Final calculation**:
Therefore,
\[ g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right) = \frac{1}{1 + \frac{1}{\sqrt{2}}} \]
To simplify:
\[ = \frac{1}{\frac{\sqrt{2} + 1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2} + 1} \]

### Final Answer:
\[ g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right) = \frac{\sqrt{2}}{\sqrt{2} + 1} \]