Problem 30 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{x^{2}}^{3} \frac{1}{1+t} d t, x>0 $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{2x}{1+x^2}\).

Step 1: Understand Leibniz's Rule

Leibniz's Rule for differentiating under the integral sign is given by:\[ \frac{d}{dx} \int_{a(x)}^{b(x)} f(t, x) \, dt = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f(t, x)}{\partial x} \, dt \]In this problem, the integral is \(\int_{x^2}^{3} \frac{1}{1+t} \, dt\) and does not explicitly depend on \(x\). Therefore, \(\partial f(t, x)/\partial x = 0\). As such, we will only focus on the first two terms of Leibniz's Rule.

Step 2: Identify the Functions and Their Derivatives

Here, \(a(x) = x^2\) and \(b(x) = 3\). The function \(f(t) = \frac{1}{1+t}\) is independent of \(x\) in this case.Derivatives:- \(a'(x) = \frac{d}{dx}(x^2) = 2x\)- \(b'(x) = \frac{d}{dx}(3) = 0\) since 3 is a constant.

Step 3: Apply Leibniz's Rule

Using Leibniz's Rule, substitute the functions and their derivatives:\[ \frac{d}{dx} \left( \int_{x^2}^{3} \frac{1}{1+t} \, dt \right) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) \]Substitute:- \(f(3, x) = \frac{1}{1+3} = \frac{1}{4}\)- \(f(x^2, x) = \frac{1}{1+x^2}\)So, the derivative is:\[ \frac{1}{4} \cdot 0 - \frac{1}{1+x^2} \cdot (2x) = -\frac{2x}{1+x^2} \]

Step 4: Present the Final Derivative

The differentiation under the integral sign, based on the steps above, results in:\[ \frac{dy}{dx} = -\frac{2x}{1+x^2} \]

Key Concepts

Differentiation Under the Integral Sign

Differentiation under the integral sign is a powerful technique that allows us to find the derivative of an integral whose limits and/or the integrand itself depends on a variable. This method is particularly useful when dealing with more complex integrals where direct differentiation isn't feasible. The formula to perform this is given by Leibniz's Rule, which provides a structured approach to handle such differentiations. In practice, the formula is:
\[ \frac{d}{dx} \int_{a(x)}^{b(x)} f(t, x) \, dt = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f(t, x)}{\partial x} \, dt\]
Each component plays a unique role:

• The terms \(f(b(x), x) \cdot b'(x)\) and \(-f(a(x), x) \cdot a'(x)\) account for the changing limits of the integral.
• When the integrand depends on the variable \(x\), \(\frac{\partial f(t, x)}{\partial x}\) allows us to incorporate changes in the function itself within the integration limits.

By substituting the relevant derivatives and values into this formula, we can evaluate the derivative of the integral.

Partial Derivatives

When dealing with functions of more than one variable, partial derivatives are a crucial concept. They allow us to measure how a function changes with respect to one variable while keeping all other variables constant. This is particularly significant in multivariable calculus and helps in analyzing situations where changes occur in multiple dimensions.
The notation \(\frac{\partial f(t, x)}{\partial x}\) represents the partial derivative of the function \(f(t, x)\) with respect to \(x\). This indicates that we are considering how the function varies as \(x\) shifts, while \(t\) remains fixed. In the context of Leibniz's Rule, this term is used to incorporate any change in the integrand that results from a change in \(x\) across the integration range.
In our specific case from the exercise, the function \(f(t) = \frac{1}{1+t}\) does not depend on \(x\), thus the partial derivative \(\frac{\partial f(t, x)}{\partial x}\) becomes 0. This simplifies the application of Leibniz's Rule significantly, as it removes the last term from the equation, letting us focus on just the changes from the limits.

Definite Integrals

Definite integrals are crucial in understanding the area under a curve or between curves over a specific interval. Unlike indefinite integrals, which represent a family of antiderivatives, definite integrals provide a specific numerical value corresponding to the integral's limits. The evaluation involves calculating the difference between the antiderivative evaluated at the upper limit and the antiderivative evaluated at the lower limit:
\[\int_{a}^{b} f(t) \, dt = F(b) - F(a)\]
This process fundamentally relies on the concept of limits of integration, which define where on the \(t\)-axis the area begins and ends. Depending on how these limits vary with respect to an external variable, the value of the definite integral will also change.
In the exercise, the integral is definite with the lower limit being \(x^2\) and the upper limit being 3. Consequently, as \(x\) changes, the area under the curve from \(x^2\) to 3 changes, influencing the output of the integral, and hence impacting the derivative \(\frac{dy}{dx}\). By using Leibniz's Rule, we handle how these limits' dependency on \(x\) affects the derivative calculation.