make function's x^2-2y^2 gradientvector on point (2,-1)

The gradient vector of a function \( f(x, y) \) is given by:

\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]

Given the function \( f(x, y) = x^2 - 2y^2 \), we need to compute the partial derivatives with respect to \( x \) and \( y \) first.

1. **Partial derivative with respect to \( x \):**

\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( x^2 - 2y^2 \right) = 2x
\]

2. **Partial derivative with respect to \( y \):**

\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( x^2 - 2y^2 \right) = -4y
\]

Thus, the gradient vector is:

\[
\nabla f(x, y) = \left( 2x, -4y \right)
\]

Now, substituting the point \( (x, y) = (2, -1) \):

\[
\nabla f(2, -1) = \left( 2(2), -4(-1) \right) = \left( 4, 4 \right)
\]

So, the gradient vector at the point \( (2, -1) \) is \( \mathbf{4\hat{i} + 4\hat{j}} \).

### Would you like more details or have any further questions?

Here are some related questions to explore:

1. What does the gradient vector represent geometrically?
2. How would the gradient vector change at different points on the surface?
3. Can we interpret the direction of the gradient in terms of steepest ascent?
4. How can we apply this concept in optimization problems?
5. What happens to the gradient vector when the point lies on a critical point?

**Tip:** The gradient vector always points in the direction of the steepest increase of a function!

make function's x^2-2y^2 gradient vector on point (2,-1)

Explore how to find the gradient vector of the function f(x, y) = x^2 - 2y^2 at the point (2, -1) with step-by-step calculations. Learn how to compute partial derivatives and apply them to determine the gradient vector, a fundamental concept in calculus. Suitable for high school students in Grades 11-12.

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