Differentiate the function.
u = 3√t + 6√t3

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 \( t \)?
2. What is the general rule

Learn how to differentiate the function u = 3√t + 6√t³ with a step-by-step solution. This problem involves differentiation of square roots and higher power functions, suitable for advanced high school students or anyone studying calculus.

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