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

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1. Rewrite the function:
   $$
   f(t)=88^{t^3}
   $$

2. Express the function in exponential form using the identity $$a^b=e^{b\ln a}$$:
   $$
   f(t)=e^{t^3\ln 88}
   $$

3. Differentiate $$ f(t) $$ using the chain rule. The derivative of $$ e^{u(t)} $$ is $$ e^{u(t)}u'(t) $$ where $$ u(t)=t^3\ln 88 $$:
   $$
   u'(t)=\ln 88\cdot \frac{d}{dt}(t^3)=\ln 88\cdot 3t^2=3t^2\ln 88
   $$
   Therefore,
   $$
   f'(t)=e^{t^3\ln 88}\cdot 3t^2\ln 88
   $$

4. Substitute back to the original form:
   $$
   f'(t)=88^{t^3}\cdot 3t^2\ln 88
   $$

Thus,  
$$
f^{\prime}(t)=3t^2\ln(88)\cdot88^{t^3}
$$
The derivative of $$ f(t) = 88^{t^3} $$ is $$ f^{\prime}(t) = 3t^2 \ln(88) \cdot 88^{t^3} $$.

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

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