Problem 25 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=\frac{\ln x}{4 y} $$

Step-by-Step Solution

Answer

The derivatives are: \( f_x = \frac{1}{4xy}, f_y = -\frac{\ln x}{4y^2}, f_{xx} = -\frac{1}{4x^2y}, f_{yy} = \frac{\ln x}{2y^3}, f_{xy} = -\frac{1}{4xy^2}, f_{yx} = -\frac{1}{4xy^2} \).

Step 1: Find the partial derivative with respect to x

The function is given by \( f(x, y) = \frac{\ln x}{4 y} \). To find \( f_x \), treat \( y \) as a constant and differentiate \( \ln x \) with respect to \( x \). The derivative of \( \ln x \) is \( \frac{1}{x} \). Therefore, \[ f_x(x, y) = \frac{1}{4y}\cdot \frac{1}{x} = \frac{1}{4xy}. \]

Step 2: Find the partial derivative with respect to y

To find \( f_y \), treat \( x \) as a constant. Differentiate \( \frac{1}{4y} \) with respect to \( y \), which gives \( -\frac{1}{4y^2} \). Thus, \[ f_y(x, y) = \ln x \cdot \left( -\frac{1}{4y^2} \right) = -\frac{\ln x}{4y^2}. \]

Step 3: Find the second partial derivative with respect to x

For \( f_{xx} \), differentiate \( f_x = \frac{1}{4xy} \) with respect to \( x \). The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Therefore, \[ f_{xx} = \frac{-1}{4xy^2}. \]

Step 4: Find the second partial derivative with respect to y

For \( f_{yy} \), differentiate \( f_y = -\frac{\ln x}{4y^2} \) with respect to \( y \). The derivative of \( \frac{1}{y^2} \) is \( -\frac{2}{y^3} \), thus: \[ f_{yy} = \ln x \cdot \frac{2}{4y^3} = \frac{\ln x}{2y^3}. \]

Step 5: Find the mixed partial derivative \(f_{xy}\)

For \( f_{xy} \), differentiate \( f_x = \frac{1}{4xy} \) with respect to \( y \). This is similar to differentiating \( \frac{1}{y} \) giving \( -\frac{1}{y^2} \), so: \[ f_{xy} = -\frac{1}{4xy^2}. \]

Step 6: Find the mixed partial derivative \(f_{yx}\)

For \( f_{yx} \), differentiate \( f_y = -\frac{\ln x}{4y^2} \) with respect to \( x \). The derivative of \( \ln x \) is \( \frac{1}{x} \), hence: \[ f_{yx} = -\frac{1}{4xy^2}. \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus that helps us differentiate functions composed of other functions. When dealing with partial derivatives, the Chain Rule simplifies the process by methodically finding derivatives of each function involved. This is particularly useful for functions with more than one variable.

Let's say you're given a function like \( f(x, y) = \frac{\ln x}{4y} \). To find the partial derivative with respect to each variable while considering others as constants applies the Chain Rule conceptually.

• For \( f_x \), treat \( y \) as a constant and differentiate \( \ln x \) with respect to \( x \). This utilizes the basic derivative property of \( \ln x \) turning into \( \frac{1}{x} \).
• For \( f_y \), treat \( x \) as a constant and differentiate \( \frac{1}{4y} \), resulting in \( -\frac{1}{4y^2} \) due to derivative rules for functions like \( y \).

Overall, the Chain Rule streamlines finding partial derivatives in multivariable calculus.

Mixed Partial Derivatives

Mixed partial derivatives involve taking the partial derivative of a function first with respect to one variable and then with respect to another.

In the context of \( f(x, y) = \frac{\ln x}{4y} \):

• Compute \( f_{xy} \) by first differentiating \( f \) with respect to \( x \), and then again with respect to \( y \).
• Similarly, \( f_{yx} \) is found by differentiating \( f \) first with respect to \( y \) and then \( x \).

Mixed partial derivatives such as \( f_{xy} \) should typically be equal to \( f_{yx} \), assuming all necessary conditions of continuity are satisfied. For our function:

• \( f_{xy} = -\frac{1}{4xy^2} \)
• \( f_{yx} = -\frac{1}{4xy^2} \)

This equivalence demonstrates Clairaut's Theorem for equality of mixed partials, stressing the importance of smoothness for real-world computations.

Second Partial Derivatives

Second partial derivatives help us analyze functions more comprehensively. This involves taking the partial derivative of a function not once, but twice, and often provides information about the function's curvature.

For \( f(x, y) = \frac{\ln x}{4y} \), proceed as follows:

• Compute \( f_{xx} \) by differentiating \( f_x \) with respect to \( x \) again, resulting in: \(-\frac{1}{4x^2y}\).
• Likewise, determine \( f_{yy} \) by differentiating \( f_y \) with respect to \( y \), leading to: \( \frac{\ln x}{2y^3}\).

Second partial derivatives can tell us a lot about the behavior of functions. They are crucial in assessing local extremities and the concavity or convexity around a point in multivariable calculus contexts.