Problem 15 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Components

The function is given as \( f(x, y) = \ln \left( \frac{x^{2} + y}{y} \right) \). First, identify \( u(x, y) = \frac{x^{2} + y}{y} \) which indicates that \( f(x, y) = \ln(u) \). We will use this form to differentiate \( f \) with respect to both \( x \) and \( y \).

Step 2: Differentiate with Respect to x

Start by differentiating the inner function \( u(x, y) \) with respect to \( x \): \( \frac{\partial }{\partial x}u(x, y) = \frac{2x}{y} \). Then, use the chain rule: \( \frac{\partial f}{\partial x} = \frac{d}{du} \ln(u) \cdot \frac{\partial u}{\partial x} = \frac{1}{u} \cdot \frac{2x}{y} \). Substitute back \( u(x, y) = \frac{x^{2} + y}{y} \) to get \( \frac{\partial f}{\partial x} = \frac{2x}{y(x^2+y)} \).

Step 3: Differentiate with Respect to y

Now, differentiate the inner function \( u(x, y) \) with respect to \( y \). Use the quotient rule: \( \frac{\partial }{\partial y}u(x, y) = \frac{y \cdot (0) - (x^2 + y) \cdot (-1)}{y^2} = \frac{x^2}{y^2} \). Again, apply the chain rule: \( \frac{\partial f}{\partial y} = \frac{1}{u} \cdot \frac{x^2}{y^2} \). Substitute back \( u(x, y) = \frac{x^{2} + y}{y} \) to get \( \frac{\partial f}{\partial y} = \frac{x^2}{y^2(x^2+y)} \).

Key Concepts

Chain Rule

The chain rule is an essential tool when you're working with composite functions. You can think of a composite function as a "function within a function." For instance, if you have a function like \( f(u(x, y)) \), where \( u \) is another function, then the chain rule helps you differentiate \( f \) by taking into account how \( u \) changes when \( x \) or \( y \) change.

In simple terms, the chain rule states:

• First, differentiate the outer function with respect to the inner function \( u \) (as if \( u \) were a standalone variable).
• Then multiply that result by the derivative of the inner function \( u \) with respect to \( x \) or \( y \).

For our function \( f(x, y) = \ln \left( \frac{x^2 + y}{y} \right) \), we saw this in action. We first found the derivative of \( \ln(u) \), which is \( \frac{1}{u} \), and then multiplied it by the derivative of \( u(x, y) = \frac{x^2 + y}{y} \). This is how we used the chain rule to find the partial derivatives of \( f \) with respect to \( x \) and \( y \).

The chain rule is very helpful in untangling complicated functions, making it simpler to find how each variable contributes to changes in the function's output.

Quotient Rule

The quotient rule is a technique used in calculus to differentiate functions that involve division. It's a bit more complex than other differentiation rules because it involves both the numerator and the denominator. But once you get the hang of it, applying it becomes straightforward.

Here's how it works:

• If you have a function \( u(x, y) = \frac{g(x, y)}{h(x, y)} \), then the quotient rule formula is:
• \( \frac{d}{dx}u = \frac{g'(x, y)h(x, y) - g(x, y)h'(x, y)}{(h(x, y))^2} \)

In our exercise, we applied the quotient rule to differentiate \( u(x, y) = \frac{x^2 + y}{y} \) with respect to \( y \). The numerator, \( g(x, y) = x^2 + y \), had a derivative of \( 1 \) with respect to \( y \). The denominator, \( h(x, y) = y \), had a derivative of \( 1 \) with respect to \( y \).

After applying the quotient rule, we obtained \( \frac{x^2}{y^2} \), which we then used with the chain rule to finalize the partial derivative with respect to \( y \). The quotient rule is incredibly useful in many real-world applications when dealing with rates and changes.

Natural Logarithm Differentiation

Differentiating functions that involve natural logarithms generally requires the knowledge of a simple rule. The natural logarithm, denoted as \( \ln \), is one of the fundamental logarithmic functions typically used in calculus.When you need to differentiate \( \ln(x) \), the rule is straightforward:

• The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \) multiplied by the derivative of \( u \) itself, \( u' \).

In the context of our original exercise, we deal with \( f(x, y) = \ln(\frac{x^2 + y}{y}) \). This is a composite function where \( u(x, y) = \frac{x^2 + y}{y} \).
The derivative of the natural logarithm \( \ln(u) \) is \( \frac{1}{u} \), and then applying the chain rule, you multiply by \( \frac{\partial u}{\partial x} \) or \( \frac{\partial u}{\partial y} \) as needed.

This process simplified finding the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). By breaking down exponential relationships into log functions, calculations become generally more manageable, which is why understanding how to differentiate natural logs is so valuable in calculus.