If `f(x)=(x)/(sqrt(1+x^(2))`, then `f(f(f(x)))` is :

Options

A. `(3x)/(sqrt(1+x^(2)))`
B. `(x)/(sqrt(1+3x^(2)))`
C. `(x^(2))/(sqrt(1+x^(2)))`
D. `(x)/(sqrt(3+3x^(2)))`

Answer

To solve the problem, we need to find \( f(f(f(x))) \) where \( f(x) = \frac{x}{\sqrt{1 + x^2}} \).

### Step 1: Calculate \( f(f(x)) \)

1. First, we need to find \( f(x) \):
\[ f(x) = \frac{x}{\sqrt{1 + x^2}} \]

2. Now, we will substitute \( f(x) \) into itself to find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right) \]
We replace \( x \) in \( f(x) \) with \( \frac{x}{\sqrt{1 + x^2}} \):
\[ f(f(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2}} \]

3. Simplifying the denominator:
\[ \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = \frac{x^2}{1 + x^2} \]
Thus,
\[ 1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = 1 + \frac{x^2}{1 + x^2} = \frac{1 + x^2 + x^2}{1 + x^2} = \frac{1 + 2x^2}{1 + x^2} \]

4. Now substituting back into \( f(f(x)) \):
\[ f(f(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{\frac{1 + 2x^2}{1 + x^2}}} = \frac{x}{\sqrt{1 + x^2}} \cdot \frac{\sqrt{1 + x^2}}{\sqrt{1 + 2x^2}} = \frac{x}{\sqrt{1 + 2x^2}} \]

### Step 2: Calculate \( f(f(f(x))) \)

1. Now we need to find \( f(f(f(x))) \):
\[ f(f(f(x))) = f\left(\frac{x}{\sqrt{1 + 2x^2}}\right) \]
Substitute \( x \) in \( f(x) \) with \( \frac{x}{\sqrt{1 + 2x^2}} \):
\[ f(f(f(x))) = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2}} \]

2. Simplifying the denominator:
\[ \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2 = \frac{x^2}{1 + 2x^2} \]
Thus,
\[ 1 + \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2 = 1 + \frac{x^2}{1 + 2x^2} = \frac{1 + 2x^2 + x^2}{1 + 2x^2} = \frac{1 + 3x^2}{1 + 2x^2} \]

3. Now substituting back into \( f(f(f(x))) \):
\[ f(f(f(x))) = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{\frac{1 + 3x^2}{1 + 2x^2}}} = \frac{x}{\sqrt{1 + 2x^2}} \cdot \frac{\sqrt{1 + 2x^2}}{\sqrt{1 + 3x^2}} = \frac{x}{\sqrt{1 + 3x^2}} \]

### Final Answer
Thus, the final result is:
\[ f(f(f(x))) = \frac{x}{\sqrt{1 + 3x^2}} \]