Problem 20 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2}^{x^{2}-2} \sqrt{3+u} d u $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 2x \sqrt{x^2 + 1} \)

Step 1: Understand Leibniz's Rule

Leibniz's Rule is used to differentiate integrals where the limits of integration are functions of the variable with respect to which we are differentiating. The rule states that if \( y = \int_{a(x)}^{b(x)} f(u) \, du \), then \( \frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \).

Step 2: Identify Functions and Their Derivatives

In the given problem, we have \( y = \int_{2}^{x^2-2} \sqrt{3+u} \, du \). Here, \( a(x) = 2 \) and \( b(x) = x^2 - 2 \). Hence, \( a'(x) = 0 \) and \( b'(x) = \frac{d}{dx}(x^2 - 2) = 2x \).

Step 3: Evaluate the Integral's Limits

The upper limit function is \( b(x) = x^2 - 2 \). The function inside the integral is \( f(u) = \sqrt{3+u} \). You will evaluate \( f(b(x)) = \sqrt{3+(x^2-2)} = \sqrt{x^2+1} \).

Step 4: Apply Leibniz's Rule

Using Leibniz's Rule: \( \frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \). Substituting, we get: \( \frac{dy}{dx} = \sqrt{x^2 + 1} \cdot 2x - \sqrt{5} \cdot 0 \).

Step 5: Simplify the Result

The result from Leibniz's Rule simplifies to \( \frac{dy}{dx} = 2x \sqrt{x^2 + 1} \). The second term is \(0\) since \( a'(x) = 0 \).

Key Concepts

Differentiation of Integrals

Differentiating integrals can be tricky, especially when the limits of integration are functions themselves. Here's where Leibniz's Rule becomes incredibly handy. It helps us differentiate an integral where the limits are not constants but functions of the variable we're focusing on. This becomes essential when solving calculus problems involving such integrals.

Leibniz's Rule states that if you have an integral of the form:

• \( y = \int_{a(x)}^{b(x)} f(u) \, du \)

Then the derivative of \( y \) with respect to \( x \) is:

• \( \frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \)

This formula essentially applies the chain rule to the limits of the integral.

In our specific exercise, the task was to differentiate the integral with limits 2 and \( x^2 - 2 \). Applying Leibniz’s Rule involved plugging in the functions \( f(u) = \sqrt{3+u} \), evaluating it at the limits, and then using their derivatives. Understanding and applying this efficiently hides within it a simpler way to tackle otherwise cumbersome problems involving multiple functions within an integral.

Upper and Lower Limits of Integration

The limits of integration define the region over which you are integrating. They can be constant, or they can be functions of the variable you are differentiating with respect to. When using Leibniz’s Rule, understanding these limits is crucial.

The problem statement provided limits \( a(x) = 2 \) for the lower limit and \( b(x) = x^2 - 2 \) for the upper limit. A key step in using Leibniz’s Rule is evaluating the derivative of these limits. For a constant lower limit, like 2, the derivative is zero. But for a functional upper limit, like \( x^2 - 2 \), the derivative is significant. Here, it is \( b'(x) = 2x \).

By understanding these derivatives, we gathered crucial information necessary to exactly apply Leibniz’s Rule to find the derivative of the integral, ensuring that each part of the derivative expression makes sense.

Calculus Problem Solving

When faced with calculus problems, knowing when and where to apply a rule or formula is vital for success. In this context, successfully applying Leibniz's Rule required several critical steps.

Firstly, identifying the problem type—an integral with variable limits—helped recognize that Leibniz's Rule was necessary. Next, differentiating the given limits and evaluating the integrand at these limits facilitated applying the rule effectively. These steps ensure that you aren't just solving a problem mechanically but understanding the underlying concepts and relationships.

In practice:

• Determine if variable limits on an integral exist and can be differentiated.
• Calculate the derivatives of these limits.
• Insert these derivatives into Leibniz’s formula.
• Simplify the result methodically.

Following this structured approach not only aids in correctly solving such problems but also deepens comprehension of calculus principles broadly. This keeps math learning fluid and productive.