for f(x,y) = 2ye^(8xy) find the derivative in terms of y

To find the derivative of \( f(x, y) = 2y e^{8xy} \) with respect to \( y \), we treat \( x \) as a constant and apply the product rule, as \( f(x, y) \) is the product of two functions of \( y \): \( 2y \) and \( e^{8xy} \).

### Step-by-step solution:

1. **Function definition:**
   \[
   f(x, y) = 2y \cdot e^{8xy}
   \]
   
2. **Product Rule:**
   The derivative of a product of two functions \( u(y) \cdot v(y) \) is given by:
   \[
   \frac{d}{dy} [u(y) \cdot v(y)] = u'(y) \cdot v(y) + u(y) \cdot v'(y)
   \]
   Here, \( u(y) = 2y \) and \( v(y) = e^{8xy} \).

3. **Derivatives of each part:**
   - Derivative of \( u(y) = 2y \) with respect to \( y \):
     \[
     u'(y) = 2
     \]
   - Derivative of \( v(y) = e^{8xy} \) with respect to \( y \):
     Use the chain rule:
     \[
     v'(y) = e^{8xy} \cdot \frac{d}{dy} (8xy) = e^{8xy} \cdot 8x
     \]

4. **Applying the product rule:**
   Now, apply the product rule to \( f(x, y) \):
   \[
   \frac{\partial}{\partial y} f(x, y) = 2 \cdot e^{8xy} + 2y \cdot (e^{8xy} \cdot 8x)
   \]
   
5. **Simplifying the expression:**
   Factor out \( e^{8xy} \):
   \[
   \frac{\partial}{\partial y} f(x, y) = 2 e^{8xy} \left( 1 + 8xy \right)
   \]

### Final result:
\[
\frac{\partial}{\partial y} f(x, y) = 2 e^{8xy} (1 + 8xy)
\]

Would you like more details or have any further questions?

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Here are five related questions:

1. How do you apply the chain rule for more complex functions?
2. What is the product rule in calculus, and when should it be used?
3. How would the result change if \( f(x, y) \) involved a more complex function of \( y \)?
4. What is the derivative of \( e^{kxy} \) in terms of \( y \), and how does it generalize?
5. How does implicit differentiation work when differentiating with respect to \( y \)?

**Tip:** When differentiating exponential functions like \( e^{u(x,y)} \), always remember to apply the chain rule if the exponent is a function of more than one variable.

Learn how to differentiate the function f(x, y) = 2y * e^(8xy) with respect to y using the product rule and chain rule. This problem involves exponential functions and partial differentiation, suitable for students in advanced high school calculus or early college-level calculus.

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