Problem 18 - Calculus for Biology and Medicine

Question

In Problems 17-24, find the indicated partial derivatives. $$ f(x, y)=x^{1 / 3} y-x y^{1 / 3} ; f_{y}(1,1) $$

Step-by-Step Solution

Answer

The partial derivative \( f_y(1,1) = \frac{2}{3} \).

Step 1: Identify the Function and Required Partial Derivative

The function given is \( f(x, y) = x^{1/3} y - x y^{1/3} \). We need to find the partial derivative of \( f \) with respect to \( y \) at the point \( (1, 1) \).

Step 2: Differentiate with Respect to y

To find \( f_y(x, y) \), take the partial derivative of the function with respect to \( y \). The function can be split into two parts: \( f(x, y) = x^{1/3} y \) and \( -x y^{1/3} \).

Step 3: Differentiate First Term

The partial derivative of \( x^{1/3} y \) with respect to \( y \) is \( x^{1/3} \), because \( x^{1/3} \) acts as a constant multiplier.

Step 4: Differentiate Second Term

For the second part, \( -x y^{1/3} \), the partial derivative with respect to \( y \) is \( -x \frac{1}{3} y^{-2/3} \) since \( d/dy( y^{1/3} ) = \frac{1}{3} y^{-2/3} \).

Step 5: Combine Partial Derivatives

Combine the derivatives from Steps 3 and 4: \( f_y(x,y) = x^{1/3} - x \frac{1}{3} y^{-2/3} \).

Step 6: Substitute the Point (1,1) into the Derivative

Substitute \( x = 1 \) and \( y = 1 \) into the expression \( f_y(x,y) = x^{1/3} - x \frac{1}{3} y^{-2/3} \). This gives \( f_y(1,1) = 1^{1/3} - 1 \cdot \frac{1}{3} \cdot 1^{-2/3} = 1 - \frac{1}{3} \).

Step 7: Simplify the Expression

Simplify \( 1 - \frac{1}{3} \) to get \( \frac{2}{3} \). Thus, \( f_y(1,1) = \frac{2}{3} \).

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics that deals with the study of change. It allows us to understand how quantities vary and evolve over time or space. One of the fundamental concepts in calculus is the derivative, which gives us the rate of change of a function. Derivatives are used extensively to solve problems involving motion, electricity, heat, sound, light, and many other phenomena. They allow us to model physical processes mathematically. In single-variable calculus, we deal with functions that have one input and one output. As we move to multivariable calculus, the complexity increases as we now have functions with multiple inputs and potentially multiple outputs. This makes understanding how each variable affects the function more intricate.

Exploring Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of more than one variable. Here, we study functions like \( f(x, y) \) where changes in both \( x \) and \( y \) influence the output. Partial derivatives come into play by allowing us to explore how the function changes with respect to one variable at a time, keeping others constant. This is essential in understanding the influence each variable exerts independently on the function. In our exercise, the function \( f(x, y) = x^{1/3} y - x y^{1/3} \) involves the variables \( x \) and \( y \). We are required to find the partial derivative with respect to \( y \), noted as \( f_y \). This shows how changes in \( y \) influence the function while treating \( x \) as a constant. Multivariable calculus finds applications in fields like engineering, physics, and economics, where systems are often dependent on multiple factors.

The Process of Derivative Computation

Derivative computation is a systematic process involving differentiation rules and theorems to find the derivative of a function. For partial derivatives, we focus on one variable at a time, applying differentiation rules similar to those used in single-variable calculus. In our specific problem, we take the following steps:

• Identify the function to differentiate: \( f(x, y) = x^{1/3} y - x y^{1/3} \).
• Differentiate each part of the function with respect to \( y \), treating \( x \) as constant.
• The derivative of \( x^{1/3} y \) with respect to \( y \) is \( x^{1/3} \).
• The derivative of \( -x y^{1/3} \) is \( -x \frac{1}{3} y^{-2/3} \).
• Combine the results to get the partial derivative: \( f_y = x^{1/3} - x \frac{1}{3} y^{-2/3} \).

Finally, evaluate this derivative at the specific point \( (1, 1) \) of interest, arriving at a numeric value of \( \frac{2}{3} \). This computation tells us the rate at which the function changes with \( y \) at that point, holding \( x \) constant.