Problem 10 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x}\) if \(\frac{x y}{x+y}=1\).

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \frac{1 - y}{x - 1} \)

Step 1: Identify Given Equation

We start by noting that the equation given is \( \frac{x y}{x+y} = 1 \). Our goal is to find the derivative \( \frac{d y}{d x} \).

Step 2: Simplify the Equation

Multiply both sides of the equation \( \frac{x y}{x+y} = 1 \) by \( x + y \) to eliminate the fraction. This gives us \( x y = x + y \).

Step 3: Rearrange Terms

Bring all terms involving \( x \) and \( y \) to one side of the equation to set it to zero: \( x y - x - y = 0 \).

Step 4: Use Implicit Differentiation

Apply implicit differentiation to the equation \( x y - x - y = 0 \) with respect to \( x \). Differentiating gives: \( \frac{d}{d x}(xy) - \frac{d}{d x}x - \frac{d}{d x}y = 0 \).

Step 5: Differentiate Each Term

Use the product rule for the term \( xy \): \( \frac{d}{d x}(xy) = y + x\frac{d y}{d x} \). For \( -x \) and \( -y \), the derivatives are \( -1 \) and \( -\frac{d y}{d x} \) respectively. Now, substitute these into the differentiated equation: \( y + x\frac{d y}{d x} - 1 - \frac{d y}{d x} = 0 \).

Step 6: Solve for \( \frac{d y}{d x} \)

Rearrange the equation \( y + x\frac{d y}{d x} - 1 - \frac{d y}{d x} = 0 \) to solve for \( \frac{d y}{d x} \). Group \( \frac{d y}{d x} \) terms: \( (x-1)\frac{d y}{d x} = 1 - y \). Then, isolate \( \frac{d y}{d x} \) to find \( \frac{d y}{d x} = \frac{1 - y}{x - 1} \).

Key Concepts

Derivative

The concept of a derivative represents how a function changes as its input changes. It gives us a way to understand the rate of change or the slope of the function at any given point.

• The derivative of a function, often denoted as \( \frac{d y}{d x} \), specifically for functions of one variable, measures this rate of change.
• In our exercise, finding the derivative involves differentiating the given implicit function \( \frac{x y}{x+y}=1 \).

When we differentiate a function implicitly, we treat \( y \) as a function of \( x \), even though it might not be expressed explicitly in terms of \( x \). This requires applying differentiation rules such as the product rule, as well as understanding how to work with both sides of an equation.

Product Rule

Whenever you need to differentiate a product of two functions, the product rule is your best friend. Applied to our problem, the product rule helps differentiate the term \( xy \).

• The product rule formula is: \( u\cdot v \Rightarrow u'v + uv' \).
• Here, \( u = x \) and \( v = y \), so their derivatives are \( u' = 1 \) and \( v' = \frac{d y}{d x} \).
• Applying the rule: \( \frac{d}{d x}(xy) = x \frac{d y}{d x} + y \).

This step is vital when working with terms that are products of variables, ensuring each part of the expression is differentiated correctly. The rest of the equation follows standard differentiation techniques, but the product rule is essential for handling products specifically.

Rearranging Equations

After obtaining a differentiated equation, you often need to rearrange it to solve for the specific term you are interested in, typically the derivative. In our exercise, we move terms around to isolate \( \frac{d y}{d x} \).

• Start by grouping all terms containing \( \frac{d y}{d x} \) and moving constants or other terms to the opposite side of the equation.
• In this example, rearrange \( y + x \frac{d y}{d x} - 1 - \frac{d y}{d x} = 0 \) to isolate \( \frac{d y}{d x} \): \( (x-1)\frac{d y}{d x} = 1 - y \).
• The final step is to solve for \( \frac{d y}{d x} \), yielding \( \frac{d y}{d x} = \frac{1 - y}{x - 1} \).

Rearrangement is crucial as it converts the differentiated equation into a more useful form, allowing you to solve for the derivative explicitly. By isolating \( \frac{d y}{d x} \), we make it possible to understand the specific rate of change of \( y \) in relation to \( x \).