A function `y=f(x)` is given by `x=1/(1+t^2)` and `y=1/(t(1+t^2))` for all `tgt0` then `f` is

Explanation

To solve the problem, we need to analyze the given functions for \( x \) and \( y \) in terms of \( t \) and determine the behavior of the function \( f(x) \) for \( t > 0 \).

### Step-by-step Solution:

1. **Given Functions**:
We have:
\[ x = \frac{1}{1 + t^2} \]
\[ y = \frac{1}{t(1 + t^2)} \]

2. **Differentiate \( x \) with respect to \( t \)**:
We will use the quotient rule for differentiation. The quotient rule states that if \( u = 1 \) and \( v = 1 + t^2 \), then:
\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
Here, \( \frac{du}{dt} = 0 \) and \( \frac{dv}{dt} = 2t \). Thus:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(0) - (1)(2t)}{(1 + t^2)^2} = \frac{-2t}{(1 + t^2)^2} \]

3. **Differentiate \( y \) with respect to \( t \)**:
Using the quotient rule again, where \( u = 1 \) and \( v = t(1 + t^2) \):
\[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
Here, \( \frac{du}{dt} = 0 \) and \( \frac{dv}{dt} = 1(1 + t^2) + t(2t) = 1 + 3t^2 \). Thus:
\[ \frac{dy}{dt} = \frac{t(1 + t^2)(0) - (1)(1 + 3t^2)}{(t(1 + t^2))^2} = \frac{-(1 + 3t^2)}{t^2(1 + t^2)^2} \]

4. **Find \( \frac{dy}{dx} \)**:
To find \( \frac{dy}{dx} \), we use the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\frac{-(1 + 3t^2)}{t^2(1 + t^2)^2}}{\frac{-2t}{(1 + t^2)^2}} = \frac{(1 + 3t^2)}{2t^3} \]

5. **Analyze the sign of \( \frac{dy}{dx} \)**:
For \( t > 0 \):
- The numerator \( (1 + 3t^2) \) is always positive.
- The denominator \( 2t^3 \) is also always positive.

Therefore, \( \frac{dy}{dx} > 0 \) for all \( t > 0 \).

6. **Conclusion**:
Since \( \frac{dy}{dx} > 0 \) for \( t > 0 \), the function \( y = f(x) \) is a monotonically increasing function for \( t > 0 \).

### Final Answer:
The function \( f(x) \) is **monotonically increasing** for \( t > 0 \).