Problem 26 - Calculus with Concepts in Calculus

Question

Let \(f(x, y)=\int_{1}^{x} P(t) d t+\int_{1}^{y} Q(t) d t\), where \(P\) and \(Q\) are continuous. Find \(f_{x}\) and \(f_{y}\).

Step-by-Step Solution

Answer

The partial derivatives are: \(f_x = P(x)\) and \(f_y = Q(y)\).

Step 1: Understand the problem

We are given a function \( f(x, y) = \int_{1}^{x} P(t) \, dt + \int_{1}^{y} Q(t) \, dt \). This function involves definite integrals with respect to single-variable functions \( P(t) \) and \( Q(t) \), and we need to find the partial derivatives \( f_x \) and \( f_y \).

Step 2: Recall the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(t) \) is an antiderivative of \( P(t) \), then \( \frac{d}{dx} \int_{1}^{x}P(t) \, dt = P(x) \). Similarly, \( \frac{d}{dy} \int_{1}^{y} Q(t) \, dt = Q(y) \).

Step 3: Find the partial derivative with respect to x

To find \( f_x \), treat \( y \) as a constant and differentiate the function with respect to \( x \):\[ f_x = \frac{\partial}{\partial x} \left( \int_{1}^{x} P(t) \, dt + \int_{1}^{y} Q(t) \, dt \right) = P(x) \]

Step 4: Find the partial derivative with respect to y

To find \( f_y \), treat \( x \) as a constant and differentiate the function with respect to \( y \):\[ f_y = \frac{\partial}{\partial y} \left( \int_{1}^{x} P(t) \, dt + \int_{1}^{y} Q(t) \, dt \right) = Q(y) \]

Step 5: Conclude results

Now we have both partial derivatives: \( f_x = P(x) \) and \( f_y = Q(y) \). The partial derivatives depend only on their respective variable, showing the function's separability.

Key Concepts

Partial Derivatives

Partial derivatives are crucial when dealing with functions of multiple variables. In this exercise, the function is given by an expression involving two definite integrals. When calculating a partial derivative like \( f_x \), we focus on one variable—here, \( x \). We treat \( y \) as a constant. This process is analogous to finding an ordinary derivative in single-variable calculus. However, it's essential to differentiate only the part of the function that contains the variable of interest. In this scenario, since \( y \) is considered constant, the integral with respect to \( y \) disappears when deriving with respect to \( x \). The same methodology applies to \( f_y \). This function doesn't depend on \( x \) in the case of the integral over \( y \). Hence, its partial derivative \( f_y \) is simply \( Q(y) \). Partial derivatives offer insights into how a function changes concerning each variable individually, holding the others constant.

Definite Integrals

Definite integrals play a vital role in the original exercise. They involve integrating a function between defined limits, here from 1 to \( x \) and 1 to \( y \). These integrals give the area under the curve specified by the function \( P(t) \) or \( Q(t) \) between these limits. The Fundamental Theorem of Calculus assists by linking definite integrals with derivatives, stating that if you take the derivative of an integral function, you retrieve the original function \( P(t) \) or \( Q(t) \). This theorem enables us to find the partial derivatives swiftly. With functions \( P(t) \) and \( Q(t) \) being continuous, integrating them and then differentiating neatly gives the result without additional complexities. Hence, the definite integral ensures a pathway from the integrals back to the original functions as partial derivatives.

Separability in Functions

Separability refers to the ability to break down a multivariable function into separate components, each solely dependent on one variable. In this function, \( f(x, y) = \int_{1}^{x} P(t) \, dt + \int_{1}^{y} Q(t) \, dt \), separability is clear because it consists of two distinct parts. One part solely integrates over \( x \), and the other over \( y \). This property simplifies the calculation of partial derivatives, as seen in the solution. Each part can be handled independently when differentiating. Thus, \( f_x \) depends only on \( x \), and \( f_y \) depends solely on \( y \). Understanding this separability is essential because it makes the analysis more straightforward, reducing the problem dimensions from two variables to two independent single-variable problems. Therefore, separability ensures simplicity in handling and understanding complex multivariable functions.