Let `f (x)= (x)/(sqrt(1+x^(2)))` then ubrace(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 evaluate the function \( f(x) = \frac{x}{\sqrt{1+x^2}} \) iteratively \( n \) times. Let's denote \( f^{(n)}(x) \) as the function \( f \) applied \( n \) times to \( x \).

### Step-by-Step Solution:

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

2. **First Iteration** \( f^{(1)}(x) \):
\[ f^{(1)}(x) = f(x) = \frac{x}{\sqrt{1+x^2}} \]

3. **Second Iteration** \( f^{(2)}(x) = f(f(x)) \):
- Substitute \( f(x) \) into itself:
\[ f^{(2)}(x) = f\left(\frac{x}{\sqrt{1+x^2}}\right) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+x^2}}\right)^2}} \]
- Simplifying the denominator:
\[ 1 + \left(\frac{x}{\sqrt{1+x^2}}\right)^2 = 1 + \frac{x^2}{1+x^2} = \frac{1+x^2+x^2}{1+x^2} = \frac{1+2x^2}{1+x^2} \]
- Therefore,
\[ f^{(2)}(x) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{\frac{1+2x^2}{1+x^2}}} = \frac{x}{\sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2}}{\sqrt{1+2x^2}} = \frac{x}{\sqrt{1+2x^2}} \]

4. **Third Iteration** \( f^{(3)}(x) = f(f^{(2)}(x)) \):
- Substitute \( f^{(2)}(x) \) into \( f \):
\[ f^{(3)}(x) = f\left(\frac{x}{\sqrt{1+2x^2}}\right) = \frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+2x^2}}\right)^2}} \]
- Simplifying the denominator:
\[ 1 + \left(\frac{x}{\sqrt{1+2x^2}}\right)^2 = 1 + \frac{x^2}{1+2x^2} = \frac{1+2x^2+x^2}{1+2x^2} = \frac{1+3x^2}{1+2x^2} \]
- Therefore,
\[ f^{(3)}(x) = \frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{\frac{1+3x^2}{1+2x^2}}} = \frac{x}{\sqrt{1+2x^2}} \cdot \frac{\sqrt{1+2x^2}}{\sqrt{1+3x^2}} = \frac{x}{\sqrt{1+3x^2}} \]

5. **General Pattern**:
- From the iterations, we can see a pattern forming:
\[ f^{(n)}(x) = \frac{x}{\sqrt{1+nx^2}} \]

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