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To find the gradient of the function \( f(x, y) = e^{2x} \sin(3y) \) at the point \( (x, y) = (2, -2) \), we need to compute the partial derivatives of \( f(x, y) \) with respect to both \( x \) and \( y \), and then evaluate those partial derivatives at the given point.

### Step 1: Compute \( \frac{\partial f}{\partial x} \)
We treat \( y \) as a constant and differentiate \( f(x, y) = e^{2x} \sin(3y) \) with respect to \( x \):
\[
\frac{\partial f}{\partial x} = \frac{d}{dx} \left( e^{2x} \sin(3y) \right)
\]
Using the chain rule for \( e^{2x} \), we get:
\[
\frac{\partial f}{\partial x} = 2e^{2x} \sin(3y)
\]

### Step 2: Compute \( \frac{\partial f}{\partial y} \)
Now we treat \( x \) as a constant and differentiate \( f(x, y) = e^{2x} \sin(3y) \) with respect to \( y \):
\[
\frac{\partial f}{\partial y} = \frac{d}{dy} \left( e^{2x} \sin(3y) \right)
\]
Using the chain rule for \( \sin(3y) \), we get:
\[
\frac{\partial f}{\partial y} = 3e^{2x} \cos(3y)
\]

### Step 3: Evaluate the partial derivatives at \( (x, y) = (2, -2) \)

#### Evaluate \( \frac{\partial f}{\partial x} \) at \( (2, -2) \):
\[
\frac{\partial f}{\partial x} (2, -2) = 2e^{2(2)} \sin(3(-2)) = 2e^4 \sin(-6)
\]
Using \( e^4 \approx 54.5982 \) and \( \sin(-6) \approx 0.2794 \):
\[
\frac{\partial f}{\partial x} (2, -2) = 2 \times 54.5982 \times (-0.2794) \approx -30.5071
\]

#### Evaluate \( \frac{\partial f}{\partial y} \) at \( (2, -2) \):
\[
\frac{\partial f}{\partial y} (2, -2) = 3e^{2(2)} \cos(3(-2)) = 3e^4 \cos(-6)
\]
Using \( \cos(-6) \approx 0.9602 \):
\[
\frac{\partial f}{\partial y} (2, -2) = 3 \times 54.5982 \times 0.9602 \approx 157.2321
\]

### Step 4: Write the gradient
The gradient \( \nabla f(2, -2) \) is given by the vector:
\[
\nabla f(2, -2) = \left( \frac{\partial f}{\partial x} (2, -2), \frac{\partial f}{\partial y} (2, -2) \right)
\]
Thus,
\[
\nabla f(2, -2) \approx (-30.5071, 157.2321)
\]

Let me know if you'd like more details or have any questions!

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Here are some related questions:
1. How is the chain rule applied in calculating gradients?
2. Why is the cosine function used in the derivative with respect to \( y \)?
3. How does the exponential term affect the gradient's magnitude?
4. What is the geometric interpretation of the gradient vector?
5. How would the gradient change if we modified the function slightly, such as by changing the coefficients in the sine or exponential terms?

**Tip:** Always double-check your trigonometric and exponential values for accuracy when dealing with gradients that require evaluating specific points.

Determine the gradient of f(x, y) = e^(2x) sin(3y) at (x, y) = (2, -2).

Learn how to compute the gradient of the multivariable function f(x, y) = e^(2x) sin(3y) at the point (2, -2). This problem demonstrates the use of partial derivatives and the chain rule in calculus. Detailed step-by-step solution provided, including the evaluation of exponential and trigonometric functions.

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