Math Problem Statement

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comment on a fait pour calculer derivee ∂^2f (x, y)/y ∂ x∂

Solution

To compute the second partial derivative 2fyx\frac{\partial^2 f}{\partial y \partial x}, or 2f(x,y)xy\frac{\partial^2 f(x, y)}{\partial x \partial y}, we follow these general steps:

1. Start with the function f(x,y)f(x, y):

Begin by identifying the function f(x,y)f(x, y). For example, the document provides the function: f(x,y)=(x+y2)2+ex22y2xyf(x, y) = (x + y - 2)^2 + e^{x^2 - 2y^2 - xy}

2. Compute the first partial derivative with respect to xx:

The first partial derivative of ff with respect to xx, denoted fx\frac{\partial f}{\partial x}, involves applying the differentiation rules to the function while treating yy as a constant: fx=2(x+y2)+(2xy)ex22y2xy.\frac{\partial f}{\partial x} = 2(x + y - 2) + (2x - y)e^{x^2 - 2y^2 - xy}.

3. Differentiate fx\frac{\partial f}{\partial x} with respect to yy:

Next, take the partial derivative of fx\frac{\partial f}{\partial x} with respect to yy. This involves differentiating each term in fx\frac{\partial f}{\partial x} while treating xx as constant: y[2(x+y2)+(2xy)ex22y2xy].\frac{\partial}{\partial y} \left[ 2(x + y - 2) + (2x - y)e^{x^2 - 2y^2 - xy} \right]. You'll need to apply the product rule to the exponential term and differentiate each term accordingly.

4. Obtain 2fyx\frac{\partial^2 f}{\partial y \partial x}:

After performing the differentiation in Step 3, the result will be 2fyx\frac{\partial^2 f}{\partial y \partial x}, the second mixed partial derivative.

Practical Application in the Document:

In the provided document, the partial derivatives for the function f(x,y)f(x, y) have already been computed at certain points. For example, at the point (2,1)(2, 1), the mixed second partial derivative is calculated as: 2fyx(2,1)=14.\frac{\partial^2 f}{\partial y \partial x}(2, 1) = -14.

This suggests that after performing the required differentiation, the specific value of the mixed partial derivative at (2,1)(2, 1) was determined as 14-14.


Would you like more details on the differentiation steps, or do you have any further questions?

Here are 5 questions to expand on this:

  1. How do we apply the chain rule when differentiating complex expressions like exponentials?
  2. What is the significance of mixed partial derivatives in multivariable calculus?
  3. How does the order of differentiation affect the result of mixed partial derivatives?
  4. What conditions ensure that 2fyx=2fxy\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}?
  5. How do partial derivatives contribute to finding the tangent plane to a surface?

Tip: Ensure clarity in applying product and chain rules when working with mixed partial derivatives, especially for functions involving exponentials or powers.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Partial Derivatives
Mixed Partial Derivatives
Chain Rule
Product Rule

Formulas

Second Partial Derivative

Theorems

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Suitable Grade Level

Advanced College Level