Problem 35 - Calculus and its applications

Question

Find \(f_{x x}, f_{x y}, f_{y x},\) and \(f_{y y}\). $$f(x, y)=e^{2 x y}$$

Step-by-Step Solution

Answer

\(f_{xx}=4y^2 e^{2xy}\), \(f_{xy}=f_{yx}=2x e^{2xy} + 4xy^2 e^{2xy}\), \(f_{yy}=4x^2 e^{2xy}\).

Step 1: Find the First Partial Derivative with respect to x

To find the derivative of \(f(x, y) = e^{2xy}\) with respect to \(x\), we apply the chain rule. The derivative of \(e^{u}\) is \(e^u\cdot u'\), where \(u=2xy\) and \(u'\) is 2y with respect to \(x\). Therefore, \(f_x = 2y \, e^{2xy}\).

Step 2: Find \(f_{xx}\)

To find the second partial derivative with respect to \(x\), denoted as \(f_{xx}\), we differentiate \(f_x = 2y \, e^{2xy}\) with respect to \(x\) again. Apply the product rule: \((uv)' = u'v + uv'\), where \(u=2y\) and \(v=e^{2xy}\). This gives \(f_{xx} = 4y^2 \, e^{2xy}\).

Step 3: Find the First Partial Derivative with respect to y

To find the derivative of \(f(x, y) = e^{2xy}\) with respect to \(y\), use the chain rule again. The derivative of \(u=2xy\) with respect to \(y\) is 2x. Therefore, \(f_y = 2x \, e^{2xy}\).

Step 4: Find \(f_{xy}\)

To find the mixed partial derivative \(f_{xy}\), differentiate \(f_x = 2y \, e^{2xy}\) with respect to \(y\). Using the product rule, where \(u=2y\) and \(v=e^{2xy}\), we get \(f_{xy} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy}\).

Step 5: Find \(f_{yx}\)

Since mixed partial derivatives are equal if the function is continuous and differentiable, \(f_{xy} = f_{yx}\). Hence, \(f_{yx} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy}\).

Step 6: Find \(f_{yy}\)

To find the second partial derivative with respect to \(y\), denoted as \(f_{yy}\), differentiate \(f_y = 2x \, e^{2xy}\) with respect to \(y\) again. The product rule gives \(f_{yy} = 4x^2 \, e^{2xy}\).

Key Concepts

Chain Rule

When dealing with partial derivatives, the chain rule is a valuable tool. It helps us in differentiating composite functions. In our exercise, we applied the chain rule to function \( f(x, y) = e^{2xy} \). Here's how it works:

• The chain rule states that to differentiate a composite function, we need to multiply the derivative of the outer function by the derivative of the inner function.
• For example, in the partial derivative of \( e^{2xy} \) with respect to \( x \), the inner function \( u = 2xy \) has a derivative \( u' = 2y \). Thus, employing the chain rule gives us \( f_x = 2y \cdot e^{2xy} \).

This approach allowed us to tackle both first and second order derivatives efficiently. It is crucial for exploring the behavior of functions with more than one variable.

Product Rule

The product rule comes into play when differentiating products of functions. When you have a function like \( f(x, y) = 2y \, e^{2xy} \), this rule helps us find derivatives accurately.

• The product rule formula is expressed as \( (uv)' = u'v + uv' \).
• In Step 2, when finding \( f_{xx} \), we considered \( u = 2y \) and \( v = e^{2xy} \). Differentiating \( u \) gives us \( u' = 0 \), since \( 2y \) is constant regarding \( x \), while the derivative of \( v \) using the chain rule results in \( 2y \, e^{2xy} \).
• This results in \( f_{xx} = 4y^2 \, e^{2xy} \).

The product rule is particularly handy when functions are multiplied together, allowing the calculation of their derivatives accurately and effectively.

Mixed Partial Derivatives

Mixed partial derivatives show how a function changes when different variables are varied sequentially. In our example, \( f_{xy} \) and \( f_{yx} \) reflect the changes with respect to \( x \) and \( y \), respectively.

• First, differentiate \( f_x \) with respect to \( y \), using the product rule, as explored here, results in \( f_{xy} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy} \).
• In the same essence, \( f_{yx} \) is obtained by differentiating \( f_y \) with respect to \( x \), which also results in \( f_{yx} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy} \).
• The equality \( f_{xy} = f_{yx} \) holds, assuming function continuity, due to Clairaut's theorem. This symmetry is critical in multivariable calculus, allowing for simplification in many cases.

Understanding mixed partial derivatives is key in analyzing multi-variable functions, offering insights into their behavior when variables are interdependent.