To find the derivative of the function \( y = x^2 \cdot \sqrt[3]{2x} \), let's first rewrite it in a more manageable form for differentiation. The expression \( \sqrt[3]{2x} \) can be written as \( (2x)^{1/3} \). Thus, the function becomes:

\[
y = x^2 \cdot (2x)^{1/3}
\]

### Step-by-Step Differentiation

1. **Rewrite the expression**:  
   \( y = x^2 \cdot (2x)^{1/3} \).

2. **Simplify the expression**:  
   Break down the expression \( (2x)^{1/3} \) as follows:

   \[
   (2x)^{1/3} = 2^{1/3} \cdot x^{1/3}
   \]

   Substituting this into the equation for \( y \), we get:

   \[
   y = x^2 \cdot 2^{1/3} \cdot x^{1/3}
   \]

3. **Combine like terms**:  
   Combine the powers of \( x \):

   \[
   y = 2^{1/3} \cdot x^{2 + 1/3}
   \]

   Converting \( 2 + \frac{1}{3} \) to a single fraction, we have:

   \[
   y = 2^{1/3} \cdot x^{\frac{6}{3} + \frac{1}{3}} = 2^{1/3} \cdot x^{7/3}
   \]

4. **Differentiate using the power rule**:  
   The power rule states that if \( y = x^n \), then \( \frac{dy}{dx} = n \cdot x^{n-1} \). Here, \( n = \frac{7}{3} \):

   \[
   \frac{dy}{dx} = 2^{1/3} \cdot \frac{7}{3} \cdot x^{\frac{7}{3} - 1}
   \]

5. **Simplify the exponent**:  
   Calculate the new exponent:

   \[
   \frac{7}{3} - 1 = \frac{7}{3} - \frac{3}{3} = \frac{4}{3}
   \]

   Therefore, the derivative is:

   \[
   \frac{dy}{dx} = 2^{1/3} \cdot \frac{7}{3} \cdot x^{4/3}
   \]

6. **Final expression**:  
   Rewrite the final derivative neatly:

   \[
   \frac{dy}{dx} = \frac{7 \cdot 2^{1/3}}{3} \cdot x^{4/3}
   \]

### Conclusion

The derivative of the function \( y = x^2 \cdot \sqrt[3]{2x} \) is:

\[
\frac{dy}{dx} = \frac{7 \cdot 2^{1/3}}{3} \cdot x^{4/3}
\]

Would you like to see a more detailed explanation of any step or have any other questions?

### Further Related Questions

1. How would you find the second derivative of \( y = x^2 \cdot \sqrt[3]{2x} \)?
2. What is the derivative of a general product of two functions?
3. How would you integrate the function \( y = x^2 \cdot \sqrt[3]{2x} \)?
4. Can you find the critical points of the function \( y = x^2 \cdot \sqrt[3]{2x} \)?
5. How does the chain rule apply when differentiating functions like \( y = x^2 \cdot \sqrt[3]{2x} \)?

#### Tip:
When differentiating products of functions, it's often helpful to first simplify the expression as much as possible to make differentiation easier.

Learn how to find the derivative of the function y = x^2 * cube root of (2x) with a detailed step-by-step solution. This problem involves differentiation using the power rule and product rule, suitable for advanced high school students and college-level calculus learners.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation