Problem 43 - Calculus for Biology and Medicine

Question

We are not always given the function of interest in explicit form.y\( is related to \)x\( by an implicit equation. Determine using implicit differentiation and the first derivative test whether \)y\( is an increasing or a decreasing function of \)x\( \)\ln y=1-\frac{y}{x}, \quad x>0, y>0$

Step-by-Step Solution

Answer

The function \(y\) is decreasing when \(x < y.\)

Step 1: Differentiate Implicitly

Start by differentiating both sides of the given equation with respect to \(x.\) The equation is \(\ln y = 1 - \frac{y}{x}.\) Differentiating gives \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(1) - \frac{d}{dx}\left(\frac{y}{x}\right).\) This becomes \(\frac{1}{y} \frac{dy}{dx} = 0 - \left(\frac{(x)(\frac{dy}{dx}) - y}{x^2}\right).\) For simplicity, focus on \(\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx} - y}{x^2}.\)

Step 2: Rewrite the Derivative Expression

Isolate \(\frac{dy}{dx}.\) Multiply through by \(y\) to clear the fraction: \[ y \cdot \frac{1}{y} \cdot \frac{dy}{dx} = -y \cdot \frac{x\frac{dy}{dx} - y}{x^2}. \] Simplifying, you get \(\frac{dy}{dx} = -\frac{y(x \cdot \frac{dy}{dx} - y)}{x^2}.\)

Step 3: Isolate \( \frac{dy}{dx} \)

Solve for \(\frac{dy}{dx}\) by isolating it. Simplifying the expression, we have \(\frac{dy}{dx} + \frac{y^2}{x^2} = \frac{y}{x}\cdot\frac{dy}{dx}.\) We can rearrange to solve: \(\frac{dy}{dx}\left(1 - \frac{y}{x}\right) = -\frac{y^2}{x^2}.\) Thus, \(\frac{dy}{dx} = \frac{-y^2/x^2}{1 - y/x} = \frac{-y^2}{x^2 - xy}.\)

Step 4: Analyze the Derivative Sign

To determine if \(y\) is increasing or decreasing, analyze the sign of \(\frac{dy}{dx}.\) Consider whether \(\frac{-y^2}{x^2 - xy} > 0\). This will occur if \(x^2 - xy < 0\), meaning \(x(x - y) < 0\). Since \(x > 0,\) \(x - y < 0\) or \(x \) is less than \( y.\)

Step 5: Conclusion Using First Derivative Test

If \(x < y,\ \frac{dy}{dx}\) is negative, which means \(y\) decreases as \(x\) increases. If \(x > y,\ \frac{dy}{dx}\) would be positive, indicating \(y\) increases with \(x.\) Thus, \(y\) is decreasing as \(x\) grows towards \(y.\)

Key Concepts

First Derivative Test

The first derivative test is a critical tool in calculus used to determine whether a function is increasing or decreasing at specific intervals.
In essence, it involves taking the first derivative of a function and examining the sign (positive or negative) over a given interval.
Here's how it works:

• If the first derivative \( \frac{dy}{dx} \) is positive on an interval, the function \( y \) is increasing over that interval.
• If \( \frac{dy}{dx} \) is negative on an interval, the function \( y \) is decreasing over that interval.
• If \( \frac{dy}{dx} \) changes sign at a point, that point could be a local maximum or minimum.

Applying the first derivative test is an essential step in the overall process of analyzing a function's behavior. It helps identify the nature of critical points and provides insights into where the function is rising or falling. In the provided exercise, we used this test to conclude whether \( y \) is increasing or decreasing with respect to \( x \). By examining the behavior of \( \frac{dy}{dx} \) through our implicitly differentiated expression, we can determine these intervals of increase and decrease.

Increasing and Decreasing Functions

Understanding whether a function is increasing or decreasing is fundamental in math, particularly in analyzing trends in data or changes over time.
Let's delve deeper into this concept by considering the derivative:

• A function \( y \) of \( x \) is increasing if \( \frac{dy}{dx} > 0 \), meaning the slope is positive.
• The function is decreasing if \( \frac{dy}{dx} < 0 \), indicating a negative slope.

In the context of the exercise, we evaluate \( \frac{dy}{dx} \) through implicit differentiation to decide when and where the function \( y = f(x) \) is on the rise or decline.
The derived function \( \frac{dy}{dx} = \frac{-y^2}{x^2 - xy} \) suggests that the behavior of \( y \) depends heavily on the relation between \( x \) and \( y \).
If \( x < y \), as determined by the inequality \( x(x-y)<0 \), \( \frac{dy}{dx} \) becomes negative, indicating that \( y \) decreases—contrarily, if \( x > y \), the function \( y \) is increasing.

Derivatives with Logs

When dealing with logarithmic functions, especially in implicit differentiation problems, understanding how to differentiate logs is key.
For a logarithmic function like \( \ln y \), the derivative is \( \frac{1}{y} \frac{dy}{dx} \), reflecting the natural properties of logs.
Here's a breakdown of how this plays out:

• Apply the chain rule when differentiating log expressions. For instance, \( \ln y \) produces \( \frac{1}{y} \frac{dy}{dx} \).
• For equations where logs are equated to other functions, consider differentiating all terms involved to find the rate of change concerning \( x \).

Additionally, in the specific solution provided, the derivative of \( \ln y = 1 - \frac{y}{x} \) was calculated by distinguishing both sides with respect to \( x \), leveraging the log derivative rule and applying implicit differentiation to account for \( y \) as a dependent variable. This methodology underscores how critical understanding derivatives of log functions are in comprehensively solving problems involving implicit differentiation and analyzing function behavior.