Problem 25 - The Calculus with Analytic Geometry

Question

In Exercises 25 and 26 , find \(f_{x}(x, y)\) and \(f_{y}(x, y)\). \(f(x, y)=\int_{x}^{y} \ln \sin t d t\)

Step-by-Step Solution

Answer

The partial derivatives are: \[f_{x}(x, y) = -\ln \sin x\] and \[f_{y}(x, y) = \ln \sin y\]

Step 1 - Understand the Problem

The task is to find the partial derivatives of the function given by the integral \[f(x, y) = \int_{x}^{y} \ln \sin t \, dt\]

Step 2 - Find the Partial Derivative with Respect to x

Use the Leibniz rule for differentiation under the integral sign. The partial derivative of the integral with respect to x is given by: \[f_{x}(x, y) = -\ln \sin x\]

Step 3 - Find the Partial Derivative with Respect to y

Similarly, use the Leibniz rule for differentiation under the integral sign. The partial derivative of the integral with respect to y is given by: \[f_{y}(x, y) = \ln \sin y\]

Key Concepts

Leibniz rule

Let's dive deeper into the Leibniz rule. This rule is a formula for differentiating an integral where the limits of integration are functions of the variable of differentiation. In simpler terms, we use the Leibniz rule when we need to take the derivative of an integral with variable limits. It's especially useful for understanding partial derivatives in functions defined by integrals.
The rule can be expressed as follows:
\[ \frac{d}{dx} \bigg( \textstyle \bigintssss_{a(x)}^{b(x)} f(t,x) \, dt \bigg) = f(b(x),x) \, b'(x) - f(a(x),x) \, a'(x) + \textstyle \bigintssss_{a(x)}^{b(x)} \frac{\textstyle \partial}{\textstyle \partial x} f(t,x) \, dt \]
In our case, the functions for the limits of integration are straightforward: they are simply \( a(x) = x \) and \( b(y) = y \). Notice that the integrand \( f(t) = \ln \sin t \) does not explicitly depend on \(x\) or \(y\), simplifying our application of the rule.
By applying the Leibniz rule, we can determine how the integral responds to changes in its integration limits. This approach is essential for solving problems in both single and multivariable calculus contexts.

Differentiation under the integral sign

Differentiation under the integral sign is a technique where you take the derivative of an integral whose bounds depend on the variable of differentiation. This is closely related to the Leibniz rule but focuses specifically on replacing the integral with its differentiated form.
Consider the integral from the original exercise,
\[ f(x, y) = \textstyle \bigintssss_{x}^{y} \ln \sin t \, dt \]
To find \( f_{x}(x, y) \), we treat \( x \) as a varying limit and differentiate the integral. According to the Leibniz rule:
\[ f_{x}(x, y) = -\ln \sin x \, (1) = -\ln \sin x \]
Similarly, for \( f_{y}(x, y) \), we see that the differentiation concerning \( y \) yields:
\[ f_{y}(x, y) = \ln \sin y \, (1) = \ln \sin y \]
Mastering this technique requires understanding how to apply differentiation rules in unconventional contexts, such as integrals with variable limits.

Multivariable calculus

Multivariable calculus extends the principles of single-variable calculus to functions of several variables. In this context, we analyze functions like \( f(x, y) \) where changes in more than one variable affect the outcome.
Key concepts include:

• Partial derivatives: Differentiating a function concerning one variable while keeping others constant.
• Gradient vectors: Representing the direction and rate of fastest increase.
• Multiple integrals: Extending integration to functions of several variables.

In the original problem, we are given a function \( f(x, y) \) defined by an integral and asked to find partial derivatives. We apply partial differentiation to analyze how \( f \) changes when \( x \) or \( y \) varies:
\[ \frac{\partial}{\partial x} \textstyle \bigintssss_{x}^{y} \ln \sin t \, dt = -\ln \sin x \]
and
\[ \frac{\partial}{\partial y} \textstyle \bigintssss_{x}^{y} \ln \sin t \, dt = \ln \sin y \]
This analysis demonstrates the interplay between single-variable integration and multivariable differentiation—an essential aspect of multivariable calculus.