Problem 128 - Calculus 3

Question

For the following exercises, calculate the partial derivatives. Let \(f(x, y)=\frac{x y}{x-y} . \quad\) Find \(\quad f_{x}(2,-2) \) 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: Find the partial derivative with respect to x

To calculate the partial derivative \(f_x(x, y)\), treat \(y\) as a constant and differentiate \(f(x, y) = \frac{xy}{x-y}\) with respect to \(x\) using the quotient rule. The quotient rule is given by \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u = xy\) and \(v = x-y\).First, find \(u' = y\) and \(v' = 1\). Thus, the partial derivative is:\[f_x(x, y) = \frac{y(x-y) - xy(1)}{(x-y)^2} = \frac{yx - y^2 - xy}{(x-y)^2} = \frac{-y^2}{(x-y)^2}.\]

Step 2: Evaluate \(f_x(2, -2)\)

Substitute \(x = 2\) and \(y = -2\) into \(f_x(x, y) = \frac{-y^2}{(x-y)^2}\):\[f_x(2, -2) = \frac{-(-2)^2}{(2 - (-2))^2} = \frac{-4}{4^2} = \frac{-4}{16} = -\frac{1}{4}.\]

Step 3: Find the partial derivative with respect to y

To calculate the partial derivative \(f_y(x, y)\), treat \(x\) as a constant and differentiate \(f(x, y) = \frac{xy}{x-y}\) with respect to \(y\) using the quotient rule. Here, \(u = xy\) and \(v = x-y\).Find \(u' = x\) and \(v' = -1\). Thus, the partial derivative is:\[f_y(x, y) = \frac{x(x-y) - xy(-1)}{(x-y)^2} = \frac{x^2 - xy + xy}{(x-y)^2} = \frac{x^2}{(x-y)^2}.\]

Step 4: Evaluate \(f_y(2, -2)\)

Substitute \(x = 2\) and \(y = -2\) into \(f_y(x, y) = \frac{x^2}{(x-y)^2}\):\[f_y(2, -2) = \frac{2^2}{(2 - (-2))^2} = \frac{4}{4^2} = \frac{4}{16} = \frac{1}{4}.\]

Key Concepts

Quotient Rule

The quotient rule is an essential tool in calculus used for finding the derivative of a function that is the ratio of two differentiable functions. If you have a function in the form of \( u(x, y) / v(x, y) \), the quotient rule helps to compute its derivative quite effortlessly. The formula is as follows:

• Let \( u(x, y) \) and \( v(x, y) \) be two differentiable functions, then:
• \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]

To apply it effectively, identify the functions \( u \) and \( v \), along with their derivatives \( u' \) and \( v' \). Substitute these into the formula for accurate results. This rule is not only used in single-variable calculus but also extends to multivariable functions, enhancing our toolkit in calculus.

Multivariable Calculus

Multivariable calculus is an extension of calculus in one dimension to higher dimensions. It involves functions that take vectors as input, rather than real numbers. When dealing with such functions, notions like curves, surfaces, and higher-dimensional analogs are essential. In these settings, we often look into partial derivatives rather than ordinary derivatives.

• Partial derivatives measure the rate of change of a function with respect to one variable, holding others constant.
• They are integral to fields like physics and engineering where functions depend on multiple variables.

In practice, for a function \( f(x, y) \), the partial derivative with respect to \( x \), \( f_x \), treats \( y \) as a constant and vice versa. This allows us to focus on how the function behaves when changing just one variable at a time.

Evaluating Derivatives

Evaluating derivatives is a practical skill in calculus enabling us to understand how a function changes at specific points. Once partial derivatives are obtained, they provide a snapshot of the function's behavior at those points. For example, consider the function \( f(x, y) = \frac{xy}{x-y} \):

• To find \( f_x(2, -2) \), treat \( y \) as a constant, use the quotient rule, and substitute specific values for \( x \) and \( y \).
• Similarly, \( f_y(x, y) \) reflects changes with \( y \) horizontal, evaluated at \( x = 2 \), \( y = -2 \).

These evaluations provide local linear approximations to the surface described by the function at those points. Understand each step and practice often to achieve proficiency.