Problem 32 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2+x^{2}}^{2} \cot t d t $$

Step-by-Step Solution

Answer

The derivative is \(-2x \cot(2 + x^2)\).

Step 1: Identify the exercise

The exercise requires us to find the derivative of the function \( y = \int_{2+x^2}^{2} \cot t \, dt \) with respect to \( x \). This is a problem involving differentiation of an integral with variable limits.

Step 2: Understand Leibniz's Rule

Leibniz's rule for differentiation under the integral sign states that if \( y = \int_{a(x)}^{b(x)} f(t, x) \, dt \), then \( \frac{dy}{dx} = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x} \, dt \). In our case, \( f(t) = \cot t \) and the integral has variable lower limit \( a(x) = 2 + x^2 \) and constant upper limit \( b(x) = 2 \).

Step 3: Apply Leibniz's Rule

For the given integral \( y = \int_{2+x^2}^{2} \cot t \, dt \), apply Leibniz's rule:1. Since the upper limit \( b(x) = 2 \) is constant, we have \( b'(x) = 0 \).2. The lower limit \( a(x) = 2 + x^2 \), hence \( a'(x) = 2x \).3. Evaluate \( f(t) \) at the limits: - \( f(2, x) = \cot(2) \) - \( f(2+x^2, x) = \cot(2+x^2) \)4. Since \( f(t) \) does not explicitly depend on \( x \), we have \( \frac{\partial f}{\partial x} = 0 \).

Step 4: Compute the derivative

Substitute into Leibniz's rule:\[ \frac{dy}{dx} = 0 \cdot \cot(2) - \cot(2 + x^2) \cdot 2x + \int_{2+x^2}^{2} 0 \, dt \]Simplify to get:\[ \frac{dy}{dx} = -2x \cdot \cot(2 + x^2) \]The derivative is given by the expression \(-2x \cot(2 + x^2)\).

Key Concepts

Differentiation of Integrals

Differentiation of integrals is often encountered when you need to find how a certain integral behaves as a function changes. This particularly arises when analyzing integrals where one or both of the limits depend on a variable. In the exercise, our primary goal is to differentiate an integral with respect to a variable, in this case, finding \( \frac{dy}{dx} \) for the function\[ y = \int_{2+x^2}^{2} \cot t \, dt. \]

Leibniz's rule comes into play here, enabling us to compute the derivative precisely. The rule specifies how to differentiate an integral that depends on a parameter. It allows us to break down the components by separately considering changes in the limits of integration and the function within the integral.

Overall, this differentiation process helps to understand how the integral value changes, thereby enlightening many applications in physics and engineering, where integrals express quantities dependent on mutable parameters.

Variable Limits of Integration

When dealing with integrals that have variable limits of integration, understanding these limits is crucial. Essentially, variable limits mean that either or both of the endpoints of the integral are expressions involving a variable. In the problem we examined, the lower limit is \( 2 + x^2 \) while the upper limit is a constant 2.

This influences the differentiation because as the variable—in this case \( x \)—changes, the bounds of the integral shift as well, affecting the value of the integral.

• The upper limit \( b(x) = 2 \) results in its derivative with respect to \( x \) being 0, effectively simplifying the process for the application of Leibniz’s rule.
• The lower limit \( a(x) = 2 + x^2 \) yields a derivative \( a'(x) = 2x \), which plays a key role in finding \( \frac{dy}{dx} \).

These adjustments in the limits significantly influence the outcome, appearing explicitly in the final expression when applying differentiation rules.

Partial Derivatives

Partial derivatives come into play when we look at how a function changes with respect to one variable while keeping others constant. They are vital when applying Leibniz’s rule if the integrand is a function of both \( t \) and \( x \).

• In this exercise, we identified the function within the integral as \( \cot t \), which doesn't explicitly depend on \( x \). Thus, the partial derivative \( \frac{\partial f}{\partial x} = 0 \).
• If the integrand depended on \( x \), this component would be non-zero, and we would need to integrate it over the interval, adding another layer of complexity.

Understanding partial derivatives is crucial for analyzing scenarios where changes in multiple variables affect functions simultaneously, offering insight into individual variable contributions.