Find the derivative of `y=(t^2-1)/(t^2+1)`.

Explanation

To find the derivative of the function \( y = \frac{t^2 - 1}{t^2 + 1} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form of \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( t \), then the derivative is given by:

\[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

### Step-by-Step Solution:

1. **Identify \( u \) and \( v \)**:
- Let \( u = t^2 - 1 \)
- Let \( v = t^2 + 1 \)

2. **Differentiate \( u \) and \( v \)**:
- The derivative of \( u \) with respect to \( t \) is:
\[ \frac{du}{dt} = \frac{d}{dt}(t^2 - 1) = 2t \]
- The derivative of \( v \) with respect to \( t \) is:
\[ \frac{dv}{dt} = \frac{d}{dt}(t^2 + 1) = 2t \]

3. **Apply the Quotient Rule**:
- Substitute \( u \), \( v \), \( \frac{du}{dt} \), and \( \frac{dv}{dt} \) into the quotient rule formula:
\[ \frac{dy}{dt} = \frac{(t^2 + 1)(2t) - (t^2 - 1)(2t)}{(t^2 + 1)^2} \]

4. **Simplify the Numerator**:
- Expand the numerator:
\[ = \frac{2t(t^2 + 1) - 2t(t^2 - 1)}{(t^2 + 1)^2} \]
- This simplifies to:
\[ = \frac{2t(t^2 + 1 - t^2 + 1)}{(t^2 + 1)^2} \]
- Combine like terms:
\[ = \frac{2t(2)}{(t^2 + 1)^2} = \frac{4t}{(t^2 + 1)^2} \]

5. **Final Result**:
- Thus, the derivative of \( y \) with respect to \( t \) is:
\[ \frac{dy}{dt} = \frac{4t}{(t^2 + 1)^2} \]