Problem 8 - College Algebra and Calculus: An Applied Approach

Question

Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ f(x, y)=\frac{x y}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

The first partial derivatives of the function \(f(x, y) = \frac{xy} {x^2 + y^2}\) with respect to \(x\) and \(y\) are: \[f'_x = \frac{y(x^2 + y^2) - 2x^2y} {{(x^2 + y^2)}^2}\] and \[f'_y = \frac{x (x^2 + y^2) - 2y^2x} {{(x^2 + y^2)}^2}\] respectively.

Step 1: Compute the Partial Derivative with Respect to \(x\)

Firstly, keep \(y\) constant and differentiate \(f(x, y)\) with respect to \(x\). This can be done using the quotient rule: \[f'(x, y) = \frac{g'(x)h(x) - g(x)h'(x)} {{[h(x)]}^2} \]. Here, \(g(x) = xy\) and \(h(x) = x^2 + y^2\). Thus, \[f'_x = \frac{y(x^2 + y^2) - 2x^2y} {{(x^2 + y^2)}^2}\]

Step 2: Compute the Partial Derivative with Respect to \(y\)

Therefore, keep \(x\) constant and differentiate \(f(x, y)\) with respect to \(y\). This similarly can be done using the quotient rule: \[f'(x, y) = \frac{g'(y)h(y) - g(y)h'(y)} {{[h(y)]}^2} \]. Here, \(g(y) = xy\) and \(h(y) = x^2 + y^2\). As a result, \[f'_y = \frac{x (x^2 + y^2) - 2y^2x} {{(x^2 + y^2)}^2}\]

Key Concepts

Calculus

Calculus is a branch of mathematics that studies continuous change and includes operations such as differentiation and integration. An essential component of calculus is the concept of a derivative, which measures how a function changes as its input changes. In the context of our exercise, we're specifically focusing on partial derivatives, which are a way to find the derivative of a function of multiple variables with respect to one variable at a time.

In the given function, \( f(x, y)=\frac{x y}{x^{2}+y^{2}} \), the goal is to find how \( f \) changes with respect to each variable, \( x \) and \( y \), independently. Finding the partial derivatives involves treating the other variable as a constant and applying derivative rules as if dealing with a single-variable function. This process reveals the slope of the function's surface in the direction of each axis (\( x \) or \( y \)), which is fundamental in understanding the geometry and behavior of multivariable functions.

Quotient Rule

The quotient rule is a technique for differentiating functions that are divided by other functions. It is often remembered by the mnemonic 'low d-high minus high d-low, square the bottom and away we go', which encapsulates the formula \( \frac{g'(x)h(x) - g(x)h'(x)}{{[h(x)]}^2} \).

In our exercise, we apply the quotient rule to find the partial derivatives of the function \( f(x, y) \). We define the numerator as \( g(x) = xy \) for the partial derivative with respect to \( x \), and the denominator as \( h(x) = x^2 + y^2 \). We then differentiate \( g \) and \( h \) with respect to \( x \), while treating \( y \) as a constant, and then apply the formula. The quotient rule makes it possible to efficiently manage the complexity of differentiating ratios, which would otherwise require a cumbersome application of the product and chain rules.

Multivariable Calculus

Multivariable calculus, also known as multivariate calculus, extends the concepts of single-variable calculus to functions of several variables. It plays a critical role in fields such as physics, engineering, economics, and more, providing tools for analyzing systems where multiple factors influence the outcome.

Partial derivatives, which we encounter in our exercise, are the building blocks of multivariable calculus. They represent the rate of change of a function in one direction, holding the other directions constant. In the example of \( f(x, y) \), we calculated how \( f \) changes in the direction of \( x \) and \( y \) independently. These partial derivatives are instrumental in exploring the gradients or slopes of surfaces in three-dimensional space, optimizing multivariate functions, and in the comprehension of vector fields. Moreover, they are foundational to other important concepts like divergence, curl, and Laplacians in vector calculus.