Find the derivative of the function. \[ f(t)=88^{t^{3}} \] \( f^{\prime}(t)=\square \)

We start with the function \[ f(t)=88^{t^3}. \] Recall that if we have a function of the form \[ g(t)=a^{h(t)}, \] its derivative is given by \[ g'(t)=a^{h(t)}\ln(a)\cdot h'(t). \] For our function, \( a=88 \) and \( h(t)=t^3 \). Now, we compute the derivative of \( h(t) \): \[ h'(t)=\frac{d}{dt}(t^3)=3t^2. \] Thus, applying the formula we get: \[ f'(t)=88^{t^3}\ln(88)\cdot 3t^2. \] Rearranging the factors, the derivative is: \[ f'(t)=3t^2\ln(88)\cdot88^{t^3}. \]