Problem 23 - Calculus for Biology and Medicine

Question

In Problems 17-24, find the indicated partial derivatives. $$ f(x, y)=\frac{x y}{x^{2}+2} ; f_{x}(-1,2) $$

Step-by-Step Solution

Answer

The partial derivative \( f_x(-1, 2) \) is \( \frac{-2}{9} \).

Step 1: Understand the Function

The function given is \( f(x, y) = \frac{xy}{x^2 + 2} \). We need to find the partial derivative of this function with respect to \( x \), and then evaluate it at the point \((-1, 2)\).

Step 2: Find the Partial Derivative with Respect to x

To find \( f_x(x, y) \), apply the quotient rule for partial derivatives: if \( u = xy \) and \( v = x^2 + 2 \), then \( f_x = \frac{v \, \partial u/\partial x - u \, \partial v/\partial x}{v^2} \). Here, \( \partial u/\partial x = y \) and \( \partial v/\partial x = 2x \). Thus, the derivative is:\[ f_x(x, y) = \frac{(x^2 + 2) \cdot y - xy \cdot 2x}{(x^2 + 2)^2} \]

Step 3: Simplify the Expression

Substitute \( u \) and \( v \) into the expression:\[ f_x(x, y) = \frac{(x^2 + 2)y - 2x^2y}{(x^2 + 2)^2} \]This simplifies to:\[ f_x(x, y) = \frac{xy(2 - x^2)}{(x^2 + 2)^2} \]

Step 4: Evaluate at the Given Point

Now, substitute \( x = -1 \) and \( y = 2 \) into the simplified expression:\[ f_x(-1, 2) = \frac{(-1) \cdot 2 (2 - (-1)^2)}{((-1)^2 + 2)^2} \]This becomes:\[ f_x(-1, 2) = \frac{-2(2 - 1)}{(1 + 2)^2} \]

Step 5: Calculate the Result

Simplify the expression further:\[ f_x(-1, 2) = \frac{-2 \times 1}{9} = \frac{-2}{9} \]Therefore, the partial derivative of \( f \) with respect to \( x \) at \( (-1, 2) \) is \( \frac{-2}{9} \).

Key Concepts

Quotient Rule in Calculus

The quotient rule is a handy formula when dealing with the derivative of a quotient of two functions. In the context of partial derivatives, it is used in a similar way to find the derivative with respect to one variable while keeping others constant.
This rule is pivotal when differentiating expressions such as \( f(x, y) = \frac{xy}{x^2 + 2} \).
To break it down:

• Identify \( u(x, y) = xy \) as the numerator.
• Identify \( v(x, y) = x^2 + 2 \) as the denominator.
• Apply the quotient rule formula for partial derivatives: \( f_x = \frac{v \cdot \frac{\partial u}{\partial x} - u \cdot \frac{\partial v}{\partial x}}{v^2} \).

Here, you'll need to calculate \( \frac{\partial u}{\partial x} \) and \( \frac{\partial v}{\partial x} \):

• \( \frac{\partial u}{\partial x} = y \)
• \( \frac{\partial v}{\partial x} = 2x \)

Substituting these into the rule gives the expression:\[ f_x(x, y) = \frac{(x^2 + 2)y - xy \cdot 2x}{(x^2 + 2)^2} \]This leads to the partial derivative for \( x \), helping to evaluate changes in \( f \) with respect to \( x \) while \( y \) remains constant.

Calculus for Biology: Real-World Applications

Calculus isn't just for mathematicians; it plays a crucial role in biology as well. Understanding the behavior of functions like \( \frac{xy}{x^2 + 2} \) can help explain biological phenomena where variables change together.
For example, in enzyme kinetics, reaction rates often depend on substrate concentration and enzyme activity levels, which can be expressed through similar rational functions.
Key insights calculus provides in biology include:

• Rate of Change: Calculus helps quantify how a biological system changes over time, which is crucial for understanding growth rates in populations or the spread of diseases.
• Optimization: Calculus assists in finding optimal solutions, such as maximum or minimum points, which can be important for resource allocation in environmental studies.
• Dynamic Systems: Many biological systems are dynamic, and calculus allows for modeling such time-dependent changes accurately.

By using calculus, biologists can predict outcomes and conduct more efficient experiments, thereby making informed decisions.

Evaluating Partial Derivatives in Practice

Evaluating derivatives, particularly partial derivatives, requires a step-by-step approach to ensure accuracy. The process involves derivative rules and simplifying expressions to find specific values for given conditions, just like in the problem \( f_x(-1, 2) \) for the function \( f(x, y) = \frac{xy}{x^2 + 2} \).
To evaluate the partial derivative after applying the quotient rule:

• First, simplify \( f_x(x, y) = \frac{xy(2 - x^2)}{(x^2 + 2)^2} \) to make calculations simpler.
• Substitute known values into the simplified expression \( f_x(-1, 2) \):
• Calculation becomes \( \frac{(-1) \cdot 2 \cdot (2 - (-1)^2)}{((-1)^2 + 2)^2} \).
• Simplify the expression to \( \frac{-2 \cdot (2 - 1)}{9} = \frac{-2}{9} \).

These steps ensure that each calculation is clear and correct, reducing the chance of errors in derivative evaluations. This approach is essential for tackling complex problems in calculus confidently.