Problem 6 - Active Calculus Multivariable

Question

Calculate all four second-order partial derivatives and check that \(f_{x y}=\) \(f_{y x}\). Assume the variables are restricted to a domain on which the function is defined. $$f(x, y)=e^{2 x y}$$ \(f_{x x}=\) ________. \(f_{y y}=\) ________. \(f_{x y}=\) ________. \(f_{y x}=\) ________.

Step-by-Step Solution

Answer

The four second-order partial derivatives of the function \(f(x, y) = e^{2xy}\) are: \(f_{xx} = 4y^2 e^{2xy}\) \(f_{yy} = 4x^2 e^{2xy}\) \(f_{xy} = 4x e^{2xy}\) \(f_{yx} = 4x e^{2xy}\) As can be seen, \(f_{xy} = f_{yx}\), verifying the mixed partials are equal.

Step 1: Calculate \(f_x\) and \(f_y\)

Find the first-order partial derivatives with respect to \(x\) and \(y\): $$f_x = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (e^{2xy})$$ $$f_y = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (e^{2xy})$$ Applying the chain rule, we get: $$f_x = 2y e^{2xy}$$ $$f_y = 2x e^{2xy}$$

Step 2: Calculate \(f_{xx}\) and \(f_{yy}\)

Now find the second-order partial derivatives \(f_{xx}\) and \(f_{yy}\) by taking the partial derivatives with respect to \(x\) and \(y\) respectively: $$f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (2y e^{2xy})$$ $$f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (2x e^{2xy})$$ Applying the product and chain rule, we get: $$f_{xx} = 4y^2 e^{2xy}$$ $$f_{yy} = 4x^2 e^{2xy}$$

Step 3: Calculate \(f_{xy}\) and \(f_{yx}\)

Now find the mixed second-order partial derivatives \(f_{xy}\) and \(f_{yx}\) by taking the partial derivatives with respect to \(y\) and \(x\) respectively: $$f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (2y e^{2xy})$$ $$f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (2x e^{2xy})$$ Applying the product and chain rule, we get: $$f_{xy} = 4x e^{2xy}$$ $$f_{yx} = 4x e^{2xy}$$

Step 4: Verify \(f_{xy} = f_{yx}\)

As we can see, the mixed second-order partial derivatives \(f_{xy}\) and \(f_{yx}\) are equal: $$f_{xy} = 4x e^{2xy} = f_{yx}$$ In summary, we have found all second-order partial derivatives: \(f_{xx} = 4y^2 e^{2xy}\) \(f_{yy} = 4x^2 e^{2xy}\) \(f_{xy} = 4x e^{2xy}\) \(f_{yx} = 4x e^{2xy}\)

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus when dealing with functions of multiple variables, like \(f(x, y)\). In simple terms, a partial derivative measures how the function changes as you alter one variable, keeping others constant.
For instance, \(\frac{\partial f}{\partial x}\) is the rate of change in \(f\) as \(x\) changes but \(y\) remains constant. Similarly, \(\frac{\partial f}{\partial y}\) explores changes with respect to \(y\).

• To calculate a partial derivative, differentiate the function with respect to the variable of interest.
• Any other variables are treated as constants during differentiation.
• Partial derivatives are commonly needed in physics, engineering, and economics, where functions often depend on multiple inputs.

These derivatives give us the "slope" of the function in the direction of the variable being considered.

Chain Rule

The chain rule is a vital technique used when differentiating composite functions.
For a multivariable function, such as \(f(x, y) = e^{2xy}\), the chain rule helps find derivatives with respect to each variable as they influence others.
Consider the partial derivative \(f_x\):

• First, identify an "outer" function, \(e^u\), and the "inner" function, \(u = 2xy\).
• The derivative of the outer function with respect to \(u\) is \(e^u\).
• Next, differentiate the inner function, \(u\), with respect to \(x\), yielding \(2y\).

Multiply these derivatives together, giving \(f_x = 2y \cdot e^{2xy}\). This process is repeated for \(f_y\), leading to a similar breakdown.
In essence, the chain rule connects distinct functional layers during differentiation.

Product Rule

The product rule is invaluable when differentiating expressions involving products of separate functions.
For example, in the process of finding second-order partial derivatives like \(f_{xx}\) or \(f_{yy}\), the function to be differentiated includes a product, such as \(2y \cdot e^{2xy}\).

• According to the product rule, if you have two functions \(u(x, y)\) and \(v(x, y)\), \((uv)' = u'v + uv'\).
• Apply this rule directly: Differentiate one component while holding the other as constant, then switch roles.

Thus, for \(f_{xx}\) in \(2y \cdot e^{2xy}\):

• Differentiating \(2y\) gives \(0\); keep \(e^{2xy}\) constant.
• Differentiating \(e^{2xy}\) involves using the chain rule, resulting in \(0 + 4y^2 e^{2xy}\) upon combination.

The product rule may seem complex at first but is straightforward with practice, particularly in calculus involving several variables.

Mixed Derivatives

Mixed derivatives are second-order derivatives taken with respect to two different variables, such as \(f_{xy}\).
The elegance of mixed derivatives lies in their symmetry: under mild conditions, mixed partial derivatives are equal, meaning \(f_{xy} = f_{yx}\). This is known as "Clairaut's Theorem."

• To compute \(f_{xy}\), differentiate \(f_x\) with respect to \(y\), holding \(x\) constant.
• Compute \(f_{yx}\) by differentiating \(f_y\) with respect to \(x\), holding \(y\) constant.

For \(f(x, y) = e^{2xy}\), you find \(f_{xy} = 4x e^{2xy}\) and \(f_{yx} = 4x e^{2xy}\).
This result aligns with the principle that, in most cases, mixed derivatives can be interchanged. Understanding mixed derivatives is crucial for exploring how variables interact in multi-variable functions.