Problem 7 - Calculus for Biology and Medicine

Question

In Problems \(1-8\), find the gradient of each function. $$ f(x, y)=\ln \left(\frac{x}{y}+\frac{y}{x}\right) $$

Step-by-Step Solution

Answer

The gradient of the function is \(\nabla f = \left( \frac{1}{u} \left( \frac{1}{y} - \frac{y}{x^2} \right), \frac{1}{u} \left( -\frac{x}{y^2} + \frac{1}{x} \right) \right)\), where \(u = \frac{x}{y} + \frac{y}{x}\).

Step 1: Understand the Gradient Concept

The gradient of a function with two variables is a vector showing the direction of the steepest ascent. It consists of the partial derivatives of the function with respect to each variable.

Step 2: Differentiate with Respect to x

To find the partial derivative of the function with respect to \(x\), use the chain rule and quotient rule as needed. Let \(u = \frac{x}{y} + \frac{y}{x}\), then \(\frac{\partial f}{\partial x} = \frac{1}{u} \times \left(\frac{\partial}{\partial x} \left( \frac{x}{y} + \frac{y}{x} \right) \right) = \frac{1}{u} \left( \frac{1}{y} - \frac{y}{x^2} \right)\).

Step 3: Differentiate with Respect to y

Now, differentiate the function with respect to \(y\). Similarly to the previous step, \(\frac{\partial f}{\partial y} = \frac{1}{u} \times \left(\frac{\partial}{\partial y} \left( \frac{x}{y} + \frac{y}{x} \right) \right) = \frac{1}{u} \left( -\frac{x}{y^2} + \frac{1}{x} \right)\).

Step 4: Write the Gradient as a Vector

Combine the partial derivatives to express the gradient. Therefore, the gradient of \(f\) is given by \( abla f = \left( \frac{1}{u} \left( \frac{1}{y} - \frac{y}{x^2} \right), \frac{1}{u} \left( -\frac{x}{y^2} + \frac{1}{x} \right) \right) \).

Key Concepts

Partial Derivatives

Partial derivatives are a crucial concept when dealing with multivariable functions. They help us understand how a function changes as we tweak one variable while keeping others constant. In essence, a partial derivative is like taking a slice through a multi-dimensional landscape, observing changes along one axis at a time.

In this exercise, when we calculate the partial derivative of the function \(f(x, y) = \ln \left(\frac{x}{y} + \frac{y}{x}\right)\) with respect to \(x\), we are isolating the changes due to variations in \(x\), assuming \(y\) remains constant. Similarly, differentiating with respect to \(y\) treats \(x\) as static, only examining changes as \(y\) varies.

This step-by-step approach ensures that we can accurately gauge the sensitivity of the function to changes in each of its inputs, ultimately allowing us to construct the gradient vector, which reveals the function's steepest path of ascent.

Chain Rule

When working with nested functions or compositions of functions, the chain rule becomes invaluable. It allows us to differentiate these complex entities by breaking them into manageable parts. Put simply, the chain rule is a formula for computing the derivative of the composition of two or more functions.

In our specific problem, we first set \(u = \frac{x}{y} + \frac{y}{x}\). Then, to differentiate \(f(x, y) = \ln(u)\) with respect to \(x\) or \(y\), we employ the chain rule. This involves differentiating \(f\) with respect to \(u\) and multiplying by the derivative of \(u\) with respect to the variable of interest. It breaks down the differentiation task into smaller, simpler pieces.

For instance, to find \(\frac{\partial f}{\partial x}\), we calculate \(\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{\partial u}{\partial x}\). This cascading approach, made possible by the chain rule, simplifies the complex derivatives into neat segments.

Quotient Rule

The quotient rule is a fundamental tool for differentiating expressions that are divisions of two functions. It's especially crucial when the functions forming the quotient have variables and operations that complicate direct differentiation.

The rule states that for a quotient \(\frac{g(x)}{h(x)}\), its derivative is given by:\[\left(\frac{g}{h}\right)' = \frac{g'h - gh'}{h^2}\]where \(g'\) and \(h'\) are the derivatives of \(g\) and \(h\), respectively.

In our exercise, it comes into play when differentiating terms like \(\frac{x}{y}\) and \(\frac{y}{x}\). To find \(\frac{\partial}{\partial x}\) or \(\frac{\partial}{\partial y}\), each part is treated as a separate application of the quotient rule. This allows us to handle each function slab within the larger expression, ensuring accuracy in the derivative's computation.

Application of the quotient rule helps in navigating through these fractions, allowing for precise and structured differentiation required to solve these types of problems.