Find `fof " if " f(t)=t//(1+t^(2))^(1//2)`.

Options

A. `1//(1+2t^(2))^(1//2)`
B. `t//(1+2t^(2))^(1//2)`
C. `(1+2t^(2))`
D. None of these

Answer

To find the function of function \( f \circ f \) where \( f(t) = \frac{t}{(1 + t^2)^{1/2}} \), we will follow these steps:

### Step 1: Write down the function
Given:
\[ f(t) = \frac{t}{(1 + t^2)^{1/2}} \]

### Step 2: Find \( f(f(t)) \)
To find \( f(f(t)) \), we need to substitute \( f(t) \) into itself. This means we will replace \( t \) in \( f(t) \) with \( f(t) \).

So, we have:
\[ f(f(t)) = f\left(\frac{t}{(1 + t^2)^{1/2}}\right) \]

### Step 3: Substitute \( f(t) \) into \( f \)
Now we substitute \( \frac{t}{(1 + t^2)^{1/2}} \) into the function \( f \):
\[ f\left(\frac{t}{(1 + t^2)^{1/2}}\right) = \frac{\frac{t}{(1 + t^2)^{1/2}}}{\left(1 + \left(\frac{t}{(1 + t^2)^{1/2}}\right)^2\right)^{1/2}} \]

### Step 4: Simplify the denominator
Now we simplify the denominator:
\[ \left(1 + \left(\frac{t}{(1 + t^2)^{1/2}}\right)^2\right)^{1/2} = \left(1 + \frac{t^2}{1 + t^2}\right)^{1/2} \]
\[ = \left(\frac{1 + t^2 + t^2}{1 + t^2}\right)^{1/2} = \left(\frac{1 + 2t^2}{1 + t^2}\right)^{1/2} \]

### Step 5: Substitute back into the function
Now substituting back into our expression for \( f(f(t)) \):
\[ f(f(t)) = \frac{\frac{t}{(1 + t^2)^{1/2}}}{\left(\frac{1 + 2t^2}{1 + t^2}\right)^{1/2}} \]

### Step 6: Simplify the expression
This can be simplified as:
\[ = \frac{t}{(1 + t^2)^{1/2}} \cdot \frac{(1 + t^2)^{1/2}}{(1 + 2t^2)^{1/2}} = \frac{t}{(1 + 2t^2)^{1/2}} \]

### Final Result
Thus, we have:
\[ f(f(t)) = \frac{t}{(1 + 2t^2)^{1/2}} \]