Find the unit vector in the direction for which the directional derivative of f(x,y)=squarer root of (7x^3y^3−7xy^2) at the point (−1,−1) is a maximum.

Submit a unit vector using exact values. Make sure to enter an answer for each of the response fields

To find the unit vector in the direction where the directional derivative of the function \( f(x, y) = \sqrt{7x^3 y^3 - 7xy^2} \) at the point \( (-1, -1) \) is a maximum, follow these steps:

### Step 1: Compute the gradient of \( f(x, y) \)
The directional derivative is maximized in the direction of the gradient. First, let's compute the gradient of the given function \( f(x, y) \).

We have:
\[
f(x, y) = \sqrt{7x^3 y^3 - 7xy^2}
\]
The gradient is given by:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]

#### Step 1.1: Compute \( \frac{\partial f}{\partial x} \)
Using the chain rule:
\[
\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{7x^3 y^3 - 7xy^2}} \cdot \left( 21x^2 y^3 - 7y^2 \right)
\]

#### Step 1.2: Compute \( \frac{\partial f}{\partial y} \)
Similarly, applying the chain rule:
\[
\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{7x^3 y^3 - 7xy^2}} \cdot \left( 21x^3 y^2 - 14xy \right)
\]

### Step 2: Evaluate the gradient at the point \( (-1, -1) \)
Now, evaluate the partial derivatives at \( (x, y) = (-1, -1) \).

#### Step 2.1: Evaluate \( \frac{\partial f}{\partial x} \) at \( (-1, -1) \)
Substitute \( x = -1 \) and \( y = -1 \) into the expression for \( \frac{\partial f}{\partial x} \):
\[
\frac{\partial f}{\partial x} \bigg|_{(-1, -1)} = \frac{1}{2\sqrt{7(-1)^3(-1)^3 - 7(-1)(-1)^2}} \cdot \left( 21(-1)^2(-1)^3 - 7(-1)^2 \right)
\]
Simplifying:
\[
\frac{\partial f}{\partial x} \bigg|_{(-1, -1)} = \frac{1}{2\sqrt{7(1) - 7(1)}} \cdot \left( -21 - 7 \right) = \frac{1}{0} \cdot (-28)
\]
Since the denominator becomes zero, \( \nabla f \) at \( (-1, -1) \) is undefined at this point.

### Conclusion:
The function \( f(x, y) = \sqrt{7x^3 y^3 - 7xy^2} \) is undefined at the point \( (-1, -1) \), making the gradient calculation invalid at that location. Therefore, there is no unit vector in the direction for the maximum directional derivative at that point.

If you have any further questions or want more clarification, feel free to ask!

### Additional Questions:
1. What is the domain of the function \( f(x, y) \) where it is defined?
2. How do we handle points where the function is undefined for optimization problems?
3. Can we find the maximum directional derivative at points where the gradient is undefined?
4. What are the general conditions for the existence of a gradient?
5. How does the choice of the point affect the behavior of the gradient?

**Tip**: Always check the domain of a function when dealing with square roots or other operations that impose restrictions on the inputs.

Find the unit vector in the direction for which the directional derivative of f(x,y)=√(7x^3y^3−7xy^2) at the point (−1,−1) is a maximum.

Discover how to find the unit vector in the direction where the directional derivative of the function f(x, y) = √(7x^3y^3 - 7xy^2) at the point (−1, −1) reaches its maximum. This solution covers key calculus concepts such as gradients, partial derivatives, and directional derivatives. Suitable for advanced high school and undergraduate students.

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