Problem 128 - Calculus Volume 3

Question

Let \(f(x, y)=\frac{x y}{x-y}\). Find \(f_{x}(2,-2)\) and \(f_{y}(2,-2)\).

Step-by-Step Solution

Answer

\(f_x(2,-2) = -\frac{1}{4}\) and \(f_y(2,-2) = \frac{1}{4}\).

Step 1: Understand the Required Partial Derivatives

To solve this problem, we need to find the partial derivatives \(f_x(a, b)\) and \(f_y(a, b)\). These are the partial derivatives of the function with respect to \(x\) and \(y\), evaluated at the point \((a, b) = (2, -2)\).

Step 2: Compute the Partial Derivative with respect to x

The partial derivative \(f_x(x, y)\) is found by differentiating the function \(f(x, y) = \frac{x y}{x-y}\) with respect to \(x\) while treating \(y\) as a constant. Using the quotient rule \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\), where \(u = xy\) and \(v = x-y\), we get:\[f_x(x, y) = \frac{(y \cdot (x-y) - xy \cdot 1)}{(x-y)^2} = \frac{(xy - y^2 - xy)}{(x-y)^2} = \frac{-y^2}{(x-y)^2}.\]Evaluate at \((x, y) = (2, -2)\):\[f_x(2, -2) = \frac{-(-2)^2}{(2 - (-2))^2} = \frac{-4}{4^2} = \frac{-4}{16} = -\frac{1}{4}.\]

Step 3: Compute the Partial Derivative with respect to y

The partial derivative \(f_y(x, y)\) is found by differentiating \(f(x, y) = \frac{x y}{x-y}\) with respect to \(y\) while treating \(x\) as a constant. Again using the quotient rule with \(u = xy\) and \(v = x-y\), we get:\[f_y(x, y) = \frac{(x\cdot(x-y) - xy(-1))}{(x-y)^2} = \frac{(x^2 - xy + xy)}{(x-y)^2} = \frac{x^2}{(x-y)^2}.\]Evaluate at \((x, y) = (2, -2)\):\[f_y(2, -2) = \frac{2^2}{(2 - (-2))^2} = \frac{4}{4^2} = \frac{4}{16} = \frac{1}{4}.\]

Key Concepts

Quotient Rule

In calculus, the quotient rule is an essential technique for differentiating functions that are the quotient of two other functions. If you have a function expressed as the quotient \( \frac{u}{v} \), then the derivative can be found using the quotient rule formula: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). This rule helps in finding the rate of change of the division of two functions.
Using the quotient rule requires knowing the derivatives of the numerator \( u \) and the denominator \( v \) separately.
In the given exercise, where \( f(x, y) = \frac{xy}{x-y} \), these parts are recognized as:

• \( u = xy \)
• \( v = x-y \)

Differentiating \( u \) and \( v \) with respect to \( x \) and \( y \) separately allows us to apply the quotient rule efficiently, calculating the partial derivatives needed to solve the problem.

Function of Two Variables

Functions of two variables involve expressions where two inputs, usually \( x \) and \( y \), control the output. These functions are often presented as \( f(x, y) \), where each pair of inputs maps to a single output value. A real-world example could be calculating temperature based on coordinates.
To find how one variable influences the function when the other remains constant, we use partial derivatives. Partial derivatives show how the function changes as one variable shifts slightly, offering insights into the behavior of multi-variable functions.

• Partial derivative with respect to \( x \): Differentiates \( f(x, y) \) only with \( x \) in mind, treating \( y \) as constant.
• Partial derivative with respect to \( y \): Focuses on changes in \( f(x, y) \) with shifts in \( y \), treating \( x \) as constant.

In the exercise, \( f(x, y) = \frac{xy}{x-y} \) represents a function where both \( x \) and \( y \) contribute to the result, allowing exploration of these dependencies.

Evaluating Derivatives at a Point

Once partial derivatives have been computed, they can be evaluated at specific points to understand the behavior of the function at those specific instances. Evaluating derivatives clarifies how the function changes exactly at given values of \( x \) and \( y \).
In the example provided, after deriving the partial derivatives \( f_x(x, y) \) and \( f_y(x, y) \), they were specifically evaluated at \( (2, -2) \). This process involves substituting \((x, y) = (2, -2)\) into the derivative expressions.

• For \( f_x(2, -2) \), evaluate the change in \( f \) with respect to \( x \) at point \( (2, -2) \), resulting in \( -\frac{1}{4} \).
• For \( f_y(2, -2) \), substitute into the derivative with respect to \( y \), yielding \( \frac{1}{4} \).

These calculations are crucial for understanding the instantaneous rate of change of the function at specific points, which may represent physical phenomena in a practical context.