If `f(t)=(1-t)/(1+t)` then the value of f'(1/t) is

Options

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

Answer

To find the value of \( f'\left(\frac{1}{t}\right) \) for the function \( f(t) = \frac{1-t}{1+t} \), we will follow these steps:

### Step 1: Find the derivative \( f'(t) \)

Given:
\[ f(t) = \frac{1-t}{1+t} \]

We will use the quotient rule to differentiate \( f(t) \). The quotient rule states that if \( f(t) = \frac{g(t)}{h(t)} \), then:
\[ f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{(h(t))^2} \]

Here, \( g(t) = 1-t \) and \( h(t) = 1+t \).

- First, we find \( g'(t) \) and \( h'(t) \):
- \( g'(t) = -1 \)
- \( h'(t) = 1 \)

Now, applying the quotient rule:
\[ f'(t) = \frac{(-1)(1+t) - (1-t)(1)}{(1+t)^2} \]

### Step 2: Simplify the expression

Now, simplify the numerator:
\[ f'(t) = \frac{-1 - t - (1 - t)}{(1+t)^2} \]
\[ = \frac{-1 - t - 1 + t}{(1+t)^2} \]
\[ = \frac{-2}{(1+t)^2} \]

So, we have:
\[ f'(t) = \frac{-2}{(1+t)^2} \]

### Step 3: Evaluate \( f'\left(\frac{1}{t}\right) \)

Now, we need to find \( f'\left(\frac{1}{t}\right) \):
\[ f'\left(\frac{1}{t}\right) = \frac{-2}{\left(1 + \frac{1}{t}\right)^2} \]

### Step 4: Simplify \( 1 + \frac{1}{t} \)

\[ 1 + \frac{1}{t} = \frac{t + 1}{t} \]

Now, substituting this back into the derivative:
\[ f'\left(\frac{1}{t}\right) = \frac{-2}{\left(\frac{t + 1}{t}\right)^2} \]
\[ = \frac{-2}{\frac{(t+1)^2}{t^2}} = \frac{-2t^2}{(t+1)^2} \]

### Final Result

Thus, the value of \( f'\left(\frac{1}{t}\right) \) is:
\[ \boxed{\frac{-2t^2}{(t+1)^2}} \]