Let `f (x)= (x)/(sqrt(1+x^(2)))` then n times(fo fo fo ......of)(x) ` is :

Options

A. `(x)/(sqrt(1+(sum _(r=1)^(n)r)x ^(2)))`
B. `(x)/(sqrt(1+(sum _(r=1)^(n)1)x ^(2)))`
C. `((x)/(sqrt(1+ x ^(2))))^(n)`
D. `(npi)/(sqrt(1+pi x ^(2)))`

Answer

To solve the problem, we need to find the n-th iteration of the function \( f(x) = \frac{x}{\sqrt{1 + x^2}} \).

### Step-by-Step Solution:

1. **Define the Function:**
\[ f(x) = \frac{x}{\sqrt{1 + x^2}} \]

2. **Calculate \( f(f(x)) \):**
We need to substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right) \]
Let \( y = f(x) = \frac{x}{\sqrt{1 + x^2}} \). Then:
\[ f(y) = \frac{y}{\sqrt{1 + y^2}} = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2}} \]

3. **Simplify the Denominator:**
Calculate \( 1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 \):
\[ 1 + \frac{x^2}{1 + x^2} = \frac{1 + x^2 + x^2}{1 + x^2} = \frac{1 + 2x^2}{1 + x^2} \]
Thus,
\[ \sqrt{1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2} = \sqrt{\frac{1 + 2x^2}{1 + x^2}} = \frac{\sqrt{1 + 2x^2}}{\sqrt{1 + x^2}} \]

4. **Substituting Back:**
Now substitute this back into \( f(f(x)) \):
\[ f(f(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\frac{\sqrt{1 + 2x^2}}{\sqrt{1 + x^2}}} = \frac{x}{\sqrt{1 + 2x^2}} \]

5. **Calculate \( f(f(f(x))) \):**
Now we compute \( f(f(f(x))) = f(f(f(x))) \):
\[ f(f(f(x))) = f\left(\frac{x}{\sqrt{1 + 2x^2}}\right) \]
Following the same steps as before, we find:
\[ f(f(f(x))) = \frac{x}{\sqrt{1 + 3x^2}} \]

6. **Generalizing:**
Continuing this pattern, we can see that:
\[ f^{(n)}(x) = \frac{x}{\sqrt{1 + nx^2}} \]
where \( f^{(n)}(x) \) denotes the n-th iteration of the function \( f \).

### Final Result:
Thus, the n-th iteration of the function \( f(x) \) is:
\[ f^{(n)}(x) = \frac{x}{\sqrt{1 + nx^2}} \]