We are given \( f(t) = \left( t + \frac{1}{t} \right)^a \), and we need to differentiate it with respect to \( t \). Using the chain rule: \[ \frac{d}{dt} \left( t + \frac{1}{t} \right)^a = a \left( t + \frac{1}{t} \right)^{a-1} \cdot \frac{d}{dt} \left( t + \frac{1}{t} \right) \] Now, differentiate \( t + \frac{1}{t} \): \[ \frac{d}{dt} \left( t + \frac{1}{t} \right) = 1 - \frac{1}{t^2} \] Thus, the derivative is: \[ \frac{d}{dt} \left( t + \frac{1}{t} \right)^a = a \left( t + \frac{1}{t} \right)^{a-1} \left( 1 - \frac{1}{t^2} \right) \]