u= (x^3)*(y^4) x=t^ 3 y= t^2   then find du/dt.

Given the function \( u = x^3 \cdot y^4 \), with \( x = t^3 \) and \( y = t^2 \), we want to find \(\frac{du}{dt}\).

### Step-by-step solution:

1. **Substitute the values of \( x \) and \( y \) into \( u \):**
   \[
   u = (t^3)^3 \cdot (t^2)^4
   \]
   Simplifying the exponents:
   \[
   u = t^9 \cdot t^8
   \]
   Combining the exponents:
   \[
   u = t^{17}
   \]

2. **Differentiate \( u = t^{17} \) with respect to \( t \):**
   Using the power rule for differentiation \( \frac{d}{dt}(t^n) = n t^{n-1} \), we get:
   \[
   \frac{du}{dt} = 17 t^{16}
   \]

Thus, the derivative \(\frac{du}{dt} = 17 t^{16}\).

Would you like a more detailed breakdown of any steps? Let me know if you have any questions.

### Related questions:
1. How do you differentiate a product of functions?
2. What is the chain rule in differentiation?
3. Can you explain the power rule for derivatives?
4. How do you differentiate composite functions like \( f(g(x)) \)?
5. What is the difference between partial and total derivatives?

### Tip:
When differentiating, always simplify the expressions first to avoid unnecessary complexity.

u = (x^3) * (y^4), x = t^3, y = t^2. Then find du/dt.

Explore how to differentiate u = (x^3) * (y^4) with respect to t, given x = t^3 and y = t^2. This problem demonstrates algebraic simplification of exponents and the application of the power rule. Suitable for students in Grades 10-12.

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