Problem 21 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$f(x, y)=\frac{x-y}{x+y}$$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = \frac{2y}{(x+y)^2} \), \( \frac{\partial f}{\partial y} = \frac{-2x}{(x+y)^2} \)

Step 1: Identify the Function

The function given is \( f(x, y) = \frac{x-y}{x+y} \). We need to find the first partial derivatives with respect to \( x \) and \( y \).

Step 2: Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant.Use the quotient rule \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \) where \( u = x-y \) and \( v = x+y \).Compute the derivatives:- \( u' = 1 \) (derivative of \( x \) with respect to \( x \))- \( v' = 1 \) (derivative of \( x \) with respect to \( x \))Apply the quotient rule:\[ \frac{\partial f}{\partial x} = \frac{(1)(x+y) - (x-y)(1)}{(x+y)^2} = \frac{x+y - x + y}{(x+y)^2} = \frac{2y}{(x+y)^2} \]

Step 3: Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant.Again, use the quotient rule \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \) with the same \( u \) and \( v \) as in the previous step.Compute the derivatives:- \( u' = -1 \) (derivative of \( -y \) with respect to \( y \))- \( v' = 1 \) (derivative of \( y \) with respect to \( y \))Apply the quotient rule:\[ \frac{\partial f}{\partial y} = \frac{(-1)(x+y) - (x-y)(1)}{(x+y)^2} = \frac{-x-y-x+y}{(x+y)^2} = \frac{-2x}{(x+y)^2} \]

Step 4: Summarize the Derivatives

The first partial derivatives of the function \( f(x, y) = \frac{x-y}{x+y} \) are:- With respect to \( x \): \( \frac{\partial f}{\partial x} = \frac{2y}{(x+y)^2} \)- With respect to \( y \): \( \frac{\partial f}{\partial y} = \frac{-2x}{(x+y)^2} \)

Key Concepts

Quotient Rule

The quotient rule is a crucial tool in calculus, especially when dealing with division between two functions. It enables us to find derivatives of expressions where one function is divided by another. This rule states: for two differentiable functions, say \( u \) and \( v \), the derivative of their quotient \( \frac{u}{v} \) is given by \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Breaking down the rule:

• \( u' \) is the derivative of \( u \).
• \( v' \) is the derivative of \( v \).
• The formula \( \frac{u'v - uv'}{v^2} \) helps compute the derivative of the quotient.

In our exercise, \( u = x-y \) and \( v = x+y \). We apply the quotient rule to find partial derivatives with respect to both \( x \) and \( y \). Always remember, this rule requires the denominator, \( v \), to not be zero.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of multiple variables, such as \( f(x, y) \). Unlike single-variable calculus, where functions have a single output for each input \( x \), multivariable functions consider several inputs which influence the output.There are unique considerations in multivariable calculus:

• **Partial Derivatives:** These derivatives focus on one variable at a time, treating other variables as constants.
• **Function Optimization:** Involves finding minima and maxima within multi-dimensional surfaces.
• **Integration Techniques:** Useful in higher dimensions or for multiple variables.

For our function \( f(x, y) = \frac{x-y}{x+y} \), we're interested in its behavior with respect to each variable independently. Multivariable calculus allows us to understand how changes in \( x \) or \( y \) affect the function's output.

Function Derivatives

Derivatives of functions are foundational in calculus. They describe how a function changes at any given point. When calculating derivatives, especially in multivariable functions like \( f(x, y) \), we take partial derivatives to assess each variable's influence.Key aspects of function derivatives include:

• **Rate of Change:** Describes how fast or slow a function changes.
• **Tangent Lines:** Derivatives provide slopes for tangents to curves, representing instantaneous rates of change.
• **Critical points:** Found by setting derivatives equal to zero; they help in finding local minima or maxima.

In this exercise, derivatives \( \frac{\partial f}{\partial x} = \frac{2y}{(x+y)^2} \) and \( \frac{\partial f}{\partial y} = \frac{-2x}{(x+y)^2} \) are partial derivatives indicating how the function changes as \( x \) or \( y \) varies independently. These derivatives are essential for understanding the function's behavior in different scenarios.