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

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To differentiate 
$$
f(t)=88^{t^3},
$$
we can rewrite the function using the exponential form:
$$
f(t)=e^{t^3\ln(88)}.
$$

**Step 1:** Differentiate the exponent $$ t^3 \ln(88) $$.

Since $$\ln(88)$$ is a constant, using the power rule:
$$
\frac{d}{dt}(t^3\ln(88)) = 3t^2\ln(88).
$$

**Step 2:** Differentiate the exponential function $$ e^{t^3\ln(88)} $$.

The derivative of $$ e^{u(t)} $$ is $$ e^{u(t)} \cdot u'(t) $$. Here, $$ u(t)=t^3\ln(88) $$, so:
$$
f'(t) = e^{t^3\ln(88)} \cdot \left( 3t^2\ln(88) \right).
$$

**Step 3:** Substitute back $$ e^{t^3\ln(88)} $$ with $$ 88^{t^3} $$.

Thus, the derivative is:
$$
f'(t)=3t^2\ln(88)\cdot 88^{t^3}.
$$

The final answer is:
$$
f'(t)=3t^2\ln(88)\cdot 88^{t^3}.
$$
The derivative of $$ f(t) = 88^{t^3} $$ is $$ f'(t) = 3t^2 \ln(88) \cdot 88^{t^3} $$.

Find the derivative of the function.

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