Problem 6 - Calculus for Biology and Medicine

Question

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\tan (x-2 y)\)

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = \sec^2(x - 2y) \) and \( \frac{\partial f}{\partial y} = -2\sec^2(x - 2y) \).

Step 1: Identify the Function's Structure

The function given is \( f(x, y) = \tan(x - 2y) \). This indicates a composite function where the angle inside the tangent function depends on both \( x \) and \( y \). We will perform partial differentiation on this composite function.

Step 2: Differentiate with Respect to x

To find \( \frac{\partial f}{\partial x} \), observe that \( f(x, y) = \tan(u) \) where \( u = x - 2y \). The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \). Thus, we apply the chain rule: \[\frac{\partial f}{\partial x} = \frac{d}{du}(\tan(u)) \cdot \frac{\partial u}{\partial x} = \sec^2(x - 2y) \cdot 1 = \sec^2(x - 2y).\]

Step 3: Differentiate with Respect to y

To find \( \frac{\partial f}{\partial y} \), apply the chain rule similarly, but this time differentiate \( u = x - 2y \) with respect to \( y \): \[\frac{\partial f}{\partial y} = \frac{d}{du}(\tan(u)) \cdot \frac{\partial u}{\partial y} = \sec^2(x - 2y) \cdot (-2) = -2\sec^2(x - 2y).\]This step involves differentiating the inner function \( -2y \), hence the multiplication by \(-2\).

Step 4: Combine the Results

Write the results from Steps 2 and 3 together. The partial derivatives are:\[\frac{\partial f}{\partial x} = \sec^2(x - 2y),\]\[\frac{\partial f}{\partial y} = -2\sec^2(x - 2y).\]These represent the rates of change of the function with respect to \( x \) and \( y \) respectively.

Key Concepts

Composite Function

A composite function is a composition of two or more functions, where the output of one function becomes the input of another. This idea can seem complex but let's break it down into simple steps to understand how it works. Think of it like a conveyor belt in a factory. The first function adds something to the input, and the second function uses that output to create the final product. In the exercise provided, the function given is \( f(x, y) = \tan(x - 2y) \). Here, \( x - 2y \) is the inner function, and \( \tan(u) \) is the outer function, where the inner function's output \( u = x - 2y \) is input into the outer function, the tangent. This is why it's called a composite function, as one function is "inside" another. Composite functions are important because they allow us to model complex relationships more simply by breaking them down into simpler functions that are easier to understand and manipulate.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. It simplifies the process by allowing us to find the derivative of the "outer" function and multiply it by the derivative of the "inner" function. Let's say you have a function \( g(f(x)) \), where \( f(x) \) is nested inside \( g(x) \). The chain rule states that the derivative of \( g(f(x)) \) can be found by differentiating \( g \) with respect to \( f \), and then multiplying it by the derivative of \( f \) with respect to \( x \). The formula can be written as:

• \( \frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x) \)

In our given exercise \( f(x, y) = \tan(x - 2y) \), applying the chain rule involves finding the derivative of \( \tan(u) = \sec^2(u) \), where \( u = x - 2y \). Then you multiply by the derivative of \( u \) with respect to the chosen variable \( x \) or \( y \). This is repeated to find each partial derivative: one for \( x \) and another for \( y \). The key is to recognize the inner and outer function relationship and apply the chain rule appropriately.

Calculus for Biology

Calculus, including concepts like composite functions and the chain rule, plays a crucial role in biology. It helps us understand how biological systems change and interact. For instance, consider modeling the growth of a population impacted by factors such as food supply or predation. Such a model can be created using composite functions, where individual effects are combined to describe the overall growth rate. By using partial derivatives, like in the given exercise, one can determine how sensitive a biological system is to changes in individual factors like resources and environmental conditions. Specific functions might relate to enzyme reaction rates, gene expression levels, or ecological interactions, where each component affects the other. Therefore, understanding and applying calculus in biology can allow us to predict how changes in one aspect of a system can ripple through and affect the entire system, driving biological research and insights further.