Problem 16 - Calculus for Biology and Medicine

Question

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\ln \left(\frac{x y}{x^{2}+x y}\right)\)

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = \frac{1}{x} - \frac{2x + y}{x^2 + xy} \), \(\frac{\partial f}{\partial y} = \frac{1}{y} - \frac{x}{x^2 + xy} \)

Step 1: Rewrite the function

The given function is \( f(x, y) = \ln\left(\frac{xy}{x^2 + xy}\right) \). To find partial derivatives, it helps to simplify or adjust the function. Since the function is a natural logarithm, utilize the property of logarithms: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).Rewrite it as: \[ f(x, y) = \ln(xy) - \ln(x^2 + xy) \]

Step 2: Find \(\frac{\partial f}{\partial x}\)

Begin by finding the partial derivative of \( f(x, y) \) with respect to \( x \).Start with \( \ln(xy) \):Using the chain rule and product rule, the derivative of \( \ln(xy) \) with respect to \( x \) is: \[ \frac{d}{dx} \ln(xy) = \frac{1}{xy} \cdot y = \frac{y}{xy} = \frac{1}{x} \]Now, find the derivative of \( \ln(x^2 + xy) \):Using the chain rule, \[ \frac{d}{dx} \ln(x^2 + xy) = \frac{1}{x^2 + xy} \cdot (2x + y) \] Subtract the two derivatives:\[ \frac{\partial f}{\partial x} = \frac{1}{x} - \frac{2x + y}{x^2 + xy} \]

Step 3: Find \(\frac{\partial f}{\partial y}\)

Next, compute the partial derivative of \( f(x, y) \) with respect to \( y \).Consider \( \ln(xy) \):Its derivative with respect to \( y \) is:\[ \frac{d}{dy} \ln(xy) = \frac{1}{xy} \cdot x = \frac{x}{xy} = \frac{1}{y} \]Now, for \( \ln(x^2 + xy) \):The derivative is:\[ \frac{d}{dy} \ln(x^2 + xy) = \frac{1}{x^2 + xy} \cdot x \]Putting them together, the partial derivative of \( f \) with respect to \( y \) is:\[ \frac{\partial f}{\partial y} = \frac{1}{y} - \frac{x}{x^2 + xy} \]

Key Concepts

Chain Rule

When dealing with functions involving composites, the chain rule becomes an indispensable tool. Essentially, the chain rule allows us to differentiate a composite function, which is a function within another function. Consider a function \( f(g(x)) \). The chain rule states that the derivative of this composite function is the derivative of \( f \) evaluated at \( g(x) \) multiplied by the derivative of \( g(x) \). Mathematically, it's expressed as:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
In our original exercise, we used the chain rule while finding the derivative of \( \ln(x^2 + xy) \) by viewing \( x^2 + xy \) as our inner function \( g(x) \). Its derivative, \( 2x + y \), multiplied with the derivative of \( \ln(u) \) with respect to \( u \) (which is \( \frac{1}{u} \)), gave us:

• \( \frac{d}{dx} \ln(x^2 + xy) = \frac{1}{x^2 + xy} \cdot (2x + y) \)

Understanding this rule can greatly simplify the process of differentiation when working with complex expressions.

Product Rule

The product rule is another essential differentiation technique, particularly useful when differentiating products of two or more functions. If you have a function \( u(x) \cdot v(x) \), the product rule states that the derivative is:
\[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]
In our exercise, the product rule was used implicitly while differentiating the term \( xy \). Let's break it down: when differentiating \( \ln(xy) \), the expression inside the logarithm involves a product of \( x \) and \( y \). By applying the product rule, and considering the constant (with respect to the variable of differentiation), we found:

• While differentiating with respect to \( x \): You treat \( y \) as constant, so \( \frac{d}{dx}(xy) = y \).
• While differentiating with respect to \( y \): \( x \) is treated as constant, giving \( \frac{d}{dy}(xy) = x \).

The product rule simplifies the process of differentiating expressions that involve multiplication, ensuring accuracy in the calculations.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used to differentiate functions of the form \( y = f(x)^g(x) \) or those involving products and quotients in a simpler manner by taking the logarithm of both sides. It turns products into sums and quotients into differences, which can often simplify the differentiation process.

In the given function \( f(x, y) = \ln\left(\frac{xy}{x^2 + xy}\right) \), we employed this technique. Firstly, we used the logarithm property:
\[ \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \]
This transformed the complex quotient inside the log into manageable subtractions of simpler logs, allowing us to differentiate term by term.

• After rearranging, each part like \( \ln(xy) \) and \( \ln(x^2 + xy) \) was tackled individually using known rules like the chain rule.

This strategy not only simplifies cumbersome functions but also provides clearer insights into their derivative structures. Logarithmic differentiation proves particularly useful when normal differentiation methods become intricate. With practice, this approach makes challenging derivatives much more accessible.