Problem 64 - Calculus and its applications

Question

find \(f_{x x} f_{x y}, f_{y x},\) and \(f_{y y}\). $$f(x, y)=\frac{x y}{x-y}$$

Step-by-Step Solution

Answer

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

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

To find the first partial derivative of \( f(x, y) = \frac{x y}{x-y} \) with respect to \( x \), apply the quotient rule: \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = xy \) and \( v = x-y \). Compute \( u' \) and \( v' \):\[ u' = y, \quad v' = 1. \] Thus, \( f_x = \frac{(y)(x-y) - (xy)(1)}{(x-y)^2} = \frac{xy - y^2 - xy}{(x-y)^2} = \frac{-y^2}{(x-y)^2}. \)

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

Find \( f_y \) by considering \( f(x, y) = \frac{xy}{x-y} \) as a function of \( y \). Use the quotient rule where \( u=x \cdot y \) and \( v=x-y \), with \( u' = x \) and \( v' = -1 \):\[ f_y = \frac{x(x-y) - xy(-1)}{(x-y)^2} = \frac{xx - xy + xy}{(x-y)^2} = \frac{x^2}{(x-y)^2}. \]

Step 3: Find Second Partial Derivative \( f_{xx} \)

To compute \( f_{xx} \), take the derivative of \( f_x = \frac{-y^2}{(x-y)^2} \) with respect to \( x \). Use the chain rule: \( f_{xx} = \frac{d}{dx} \left( \frac{-y^2}{(x-y)^2} \right) = -y^2 \cdot \frac{d}{dx} \left( (x-y)^{-2} \right). \) Applying derivative rules: \[ \frac{d}{dx} \left( (x-y)^{-2} \right) = -2(x-y)^{-3}. \] Thus, \( f_{xx} = 2y^2(x-y)^{-3} = \frac{2y^2}{(x-y)^3}. \)

Step 4: Find Second Partial Derivative \( f_{xy} \)

To find \( f_{xy} \), take the derivative of \( f_x = \frac{-y^2}{(x-y)^2} \) with respect to \( y \). Since \( f_x = \frac{-y^2}{(x-y)^2} \), apply the product and chain rules to get:\[ f_{xy} = \frac{d}{dy} \left( \frac{-y^2}{(x-y)^2} \right) = -2y(x-y)^{-2}. \] Thus, \( f_{xy} = \frac{-2y(x-y)^{-2}}{1} = \frac{-2y}{(x-y)^2}. \)

Step 5: Find Second Partial Derivative \( f_{yx} \)

By equality of mixed partials (if enough conditions are satisfied), \( f_{xy} = f_{yx} \). We have already computed \( f_{xy} = \frac{-2y}{(x-y)^2} \), so \( f_{yx} = \frac{-2y}{(x-y)^2}. \)

Step 6: Find Second Partial Derivative \( f_{yy} \)

To find \( f_{yy} \), differentiate \( f_y = \frac{x^2}{(x-y)^2} \) with respect to \( y \). Apply the chain rule to get \( f_{yy} = x^2 \cdot \frac{d}{dy} \left( (x-y)^{-2} \right) = x^2 \cdot (-2)(x-y)^{-3}. \) Hence, \( f_{yy} = \frac{-2x^2}{(x-y)^3}. \)

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool for differentiating ratios of functions. It's especially useful when dealing with complex expressions that meet the criteria of having a numerator and a denominator. In simple terms, if you have a function of the form \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \), the derivative \( \frac{d}{dx}\left( \frac{u}{v} \right) \) is given by the formula:

• \( \frac{u'v - uv'}{v^2} \)

In this formula:

• \( u' \) represents the derivative of \( u \)
• \( v' \) represents the derivative of \( v \)
• The numerator consists of the product of the derivative of the numerator with the denominator minus the product of the numerator with the derivative of the denominator
• The entire expression is divided by the square of the denominator

This rule helps maintain accuracy when differentiating any quotient of functions. Using the quotient rule simplifies obtaining partial derivatives with respect to one variable when functions are expressed as fractions, as shown in the first steps for finding \( f_x \) and \( f_y \).

Second Partial Derivative

Finding second partial derivatives is crucial in understanding how functions change. After obtaining the first derivative with respect to a variable, you can find the second derivative by differentiating the first derivative again.
Let's explore this concept with examples from the exercise:

• To find \( f_{xx} \), you take the derivative of \( f_x \) with respect to \( x \), yielding \( \frac{2y^2}{(x-y)^3} \).
• The mixed partial derivatives, \( f_{xy} \) and \( f_{yx} \), involve differentiating the result of the first derivative regarding both variables in either order, often using mixed derivative equality, i.e., \( f_{xy} = f_{yx} \) assuming conditions hold.
• Finally, \( f_{yy} \) can be obtained by differentiating \( f_y \) with respect to \( y \).

Second partial derivatives are vital in various applications, such as determining local maxima and minima, and often require applying rules like the chain rule (explained next) to efficiently compute these derivatives with respect to more complex compositions of functions.

Chain Rule

The chain rule links changing rates of composite functions by allowing you to differentiate functions within functions. It's a key technique when you're dealing with nested functions or when functions depend on each other in a sequence.
In the example given, chain rule applications show up in the process of working with functions in forms like \( (x-y)^{-2} \).

• If you have a function \( y = g(h(x)) \), the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \).
• This means you take the derivative of the outer function \( g \) with respect to the inner function \( h \), then multiply it by the derivative of the inner function \( h \) with respect to \( x \).

In the context of the exercise solution, you can see examples of the chain rule in steps that involve changing variables like \( (x-y)^{-2} \) to compute multiple derivatives over different terms uniquely dependent on \( y \) and \( x \). The chain rule is essential for accurately deriving these complicated expressions, ensuring correct differentiation through successive layers of a function's structure.