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



To differentiate the function \( u = 3\sqrt{t} + 6\sqrt{t^3} \), we first need to express the square roots in terms of exponents. The function can be rewritten as:

\[
u = 3t^{\frac{1}{2}} + 6t^{\frac{3}{2}}
\]

Now, differentiate each term with respect to \( t \).

### Step 1: Differentiate the first term
\[
\frac{d}{dt}(3t^{\frac{1}{2}}) = 3 \cdot \frac{1}{2} \cdot t^{\frac{1}{2} - 1} = \frac{3}{2} t^{-\frac{1}{2}}
\]

### Step 2: Differentiate the second term
\[
\frac{d}{dt}(6t^{\frac{3}{2}}) = 6 \cdot \frac{3}{2} \cdot t^{\frac{3}{2} - 1} = 9t^{\frac{1}{2}}
\]

### Final Answer:
The derivative of the function \( u \) with respect to \( t \) is:

\[
\frac{du}{dt} = \frac{3}{2} t^{-\frac{1}{2}} + 9t^{\frac{1}{2}}
\]

This can also be written as:

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

Would you like more details, or do you have any questions?

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Here are five related questions:
1. How do you differentiate a function involving a cube root?
2. What is the chain rule, and how is it used in differentiation?
3. How do you differentiate a product of two functions?
4. Can you apply differentiation to trigonometric functions?
5. What are the applications of differentiation in physics?

**Tip:** When differentiating powers of \( t \), remember that the exponent rule is key: \(\frac{d}{dt} \left( t^n \right) = n \cdot t^{n-1}\).

Learn how to differentiate the function u = 3√t + 6√t^3 with a detailed step-by-step solution. This problem covers differentiation of square roots and the application of the chain rule. Suitable for students in Grades 11-12.

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