Let's differentiate the function $u = 3\sqrt{t} + 6\sqrt{t^3}$ with respect to $t$.

First, rewrite the square root expressions using exponents:

$u = 3t^{1/2} + 6t^{3/2}$

Now, differentiate each term with respect to $t$:

1. Differentiate $3t^{1/2}$:

$\frac{d}{dt}(3t^{1/2}) = 3 \cdot \frac{1}{2}t^{-1/2} = \frac{3}{2}t^{-1/2} = \frac{3}{2\sqrt{t}}$

2. Differentiate $6t^{3/2}$:

$\frac{d}{dt}(6t^{3/2}) = 6 \cdot \frac{3}{2}t^{1/2} = 9t^{1/2} = 9\sqrt{t}$

Finally, combine the results:

$\frac{du}{dt} = \frac{3}{2\sqrt{t}} + 9\sqrt{t}$

So, the derivative of the function is:

$\boxed{\frac{du}{dt} = \frac{3}{2\sqrt{t}} + 9\sqrt{t}}$

Would you like more details or have any questions?

Here are 5 related questions you might be interested in:

1. How do you differentiate a function involving higher powers of $t$?
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