Problem 29 - Calculus with Concepts in Calculus

Question

Find \(f_{x y}\) and \(f_{y x}\) $$ f(x, y)=3 x^{2}-\sqrt{2} x y^{2}+y^{5}-2 $$

Step-by-Step Solution

Answer

Both \(f_{xy}\) and \(f_{yx}\) are \(-2\sqrt{2}y\).

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

To find \(f_x\), we must differentiate \(f(x, y) = 3x^2 - \sqrt{2}xy^2 + y^5 - 2\) with respect to \(x\). Treat \(y\) as a constant.\[ f_x = \frac{\partial}{\partial x}(3x^2 - \sqrt{2}xy^2 + y^5 - 2) \]Differentiating term by term gives:\[ f_x = 6x - \sqrt{2}y^2 \]

Step 2: Find the First Partial Derivative with respect to y

To find \(f_y\), differentiate \(f(x, y) = 3x^2 - \sqrt{2}xy^2 + y^5 - 2\) with respect to \(y\). Treat \(x\) as a constant.\[ f_y = \frac{\partial}{\partial y}(3x^2 - \sqrt{2}xy^2 + y^5 - 2) \]Differentiating term by term gives:\[ f_y = -2\sqrt{2}xy + 5y^4 \]

Step 3: Find the Second Mixed Partial Derivative \(f_{xy}\)

Now differentiate the expression for \(f_x = 6x - \sqrt{2}y^2\) with respect to \(y\).\[ f_{xy} = \frac{\partial}{\partial y}(6x - \sqrt{2}y^2) \]Calculate the derivative:\[ f_{xy} = -2\sqrt{2}y \]

Step 4: Find the Second Mixed Partial Derivative \(f_{yx}\)

Differentiate the expression for \(f_y = -2\sqrt{2}xy + 5y^4\) with respect to \(x\).\[ f_{yx} = \frac{\partial}{\partial x}(-2\sqrt{2}xy + 5y^4) \]Calculate the derivative:\[ f_{yx} = -2\sqrt{2}y \]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. When you're dealing with functions of multiple variables, like \( f(x, y) \), each variable can change independently of the others. The partial derivative of a function with respect to one variable gives us the rate at which the function changes as that variable changes, while the other variables are held constant. To compute a partial derivative, you treat the variable of interest as the usual variable in differentiation, while treating all other variables as constants.

• For example, in the function \( f(x, y) = 3x^2 - \sqrt{2}xy^2 + y^5 - 2 \), computing \( f_x \) means differentiating with respect to \( x \), and considering \( y \) as a constant. This results in \( f_x = 6x - \sqrt{2}y^2 \).
• Similarly, to find \( f_y \), we differentiate with respect to \( y \), treating \( x \) as a constant. This gives \( f_y = -2\sqrt{2}xy + 5y^4 \).

Understanding partial derivatives is crucial because it lays the groundwork for more complex concepts like gradient vectors and optimization in higher dimensions.

Mixed Partial Derivatives

Mixed partial derivatives involve taking partial derivatives with respect to multiple variables, one after the other. For a function \( f(x, y) \), mixed partial derivatives like \( f_{xy} \) and \( f_{yx} \) represent how the function changes first with respect to \( x \) and then \( y \), or vice versa.The order in which the derivatives are taken can initially seem to matter, but a fascinating result called Clairaut's Theorem tells us that, under certain conditions of continuity, the mixed partial derivatives are equal: \( f_{xy} = f_{yx} \).

• In the original exercise, we have \( f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = -2\sqrt{2}y \).
• Calculating \( f_{yx} \), or \( \frac{\partial^2 f}{\partial x \partial y} \), also yields \( -2\sqrt{2}y \), showcasing this equality.

Mixed partial derivatives are crucial for analyzing curvature and understanding complex surfaces, and they play a key role in solving partial differential equations.

Differentiation

Differentiation in calculus refers to the process of finding a derivative, which measures how a function changes as its input changes. In the context of multivariable calculus, differentiation extends beyond simple one-variable functions to include partial derivatives for multi-variable functions. This allows us to explore how a function evolves as each variable changes.Differentiation techniques remain similar to single-variable calculus but adapted for multiple variables. Here are some key elements to understand:

• Identify the variable with respect to which you're differentiating. Treat other variables as constants during this process.
• Apply standard differentiation rules, such as the power rule or product rule, to each term independently.

In the exercise given, differentiation is employed to compute both the first order partial derivatives \( f_x \) and \( f_y \), and the mixed partial derivatives \( f_{xy} \) and \( f_{yx} \). Mastering these differentiation steps is crucial, as they form the building blocks for solving complex real-world problems involving changes in multiple dimensions simultaneously.