Problem 30 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to differentiate an integral with respect to its variable limits, in this case, the lower limit is a function of the differentiation variable (\(x^2\)) and the upper limit is constant (3). This is a situation where Leibniz's Rule applies.

Step 2: Applying Leibniz's Rule

Leibniz's Rule for differentiation under the integral sign states that if you have an integral of the form \[ y = \int_{a(x)}^{b(x)} f(t, x) \, dt \]then the derivative of \(y\) with respect to \(x\) is given by:\[ \frac{dy}{dx} = f(b(x), x)b'(x) - f(a(x), x)a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \] Since\( f(t, x) = \frac{1}{1+t} \) and there is no explicit \(x\) dependence in the integrand, \(\frac{\partial}{\partial x} f(t,x) = 0\). So, Leibniz's formula simplifies to:\[ \frac{dy}{dx} = f(b(x), x)b'(x) - f(a(x), x)a'(x) \]

Step 3: Computing Components of Leibniz's Rule

In our problem:\( a(x) = x^2 \), \( b(x) = 3 \), and \( f(t) = \frac{1}{1+t} \). \( \begin{align*} b'(x) &= 0, \ a'(x) &= 2x . \end{align*} \)Since \(b(x) = 3\), the first term of the formula \(f(3, x)b'(x)\) is zero because \(b'(x) = 0\). The second term \(f(x^2, x) a'(x)\) becomes \(f(x^2) imes 2x = \frac{1}{1 + x^2} imes 2x\).

Step 4: Combining the Results

Thus, using the simplified form of Leibniz's Rule, we have:\[ \frac{dy}{dx} = 0 - \left(\frac{1}{1 + x^2} \right) (2x) = -\frac{2x}{1+x^2} \] This result is the derivative of the given integral with respect to \(x\).

Key Concepts

Differentiation under the Integral Sign

Differentiation under the integral sign is a technique that allows us to differentiate an integral whose limits or the integrand are functions of a variable, say \(x\). This is incredibly useful when dealing with integrals where the limits or the function inside is changing with respect to \(x\). In such cases, you can't simply treat the derivative as a regular constant case.A vital tool in this technique is **Leibniz's Rule**. It provides a formula to find the derivative of an integral with respect to the variable without solving the integral directly. The general form of Leibniz's Rule is:

• \( f(b(x), x)b'(x) \) : Adjusts for changes in the upper limit.
• \( - f(a(x), x)a'(x) \) : Adjusts for changes in the lower limit.
• \( + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \) : Accounts for any direct change in the function inside the integral.

This method can seem quite abstract at first, but it becomes intuitive once broken down into these manageable pieces. The integral solution is adjusted based on each part of the rule, allowing us to understand how each component participates in the overall differentiation process.

Variable Limits of Integration

Variable limits in integration significantly impact how we calculate the derivative of an integral function. In this context, the **limits of integration** are functions themselves, dependent on a variable, often denoted as \(x\). For example, in the problem given, the lower limit is \(x^2\), while the upper is a constant \(3\).When differentiating such integrals, we must consider how each limit moves and influences the result. The rule of thumb is simple:

• When the upper limit is a function of \(x\), the derivative depends on the rate of change of this limit, \(b'(x)\).
• When the lower limit varies with \(x\), its influence is captured by \(-a'(x)\), a key component quantified by the negative in the Leibniz's formula.

In our specific example, since the upper limit is constant, its derivative is zero. This simplifies calculations as we only need to consider the effect of the changing lower limit, \(x^2\), and its derivative, \(2x\). This insight allows for more efficient mathematical maneuvering.

Partial Derivatives

Partial derivatives play a crucial role when differentiating functions with multiple variables, even when using Leibniz's rule. In the context of differentiation under the integral sign, they particularly help in understanding how the integrand might change directly with respect to \(x\).The notion of the partial derivative \(\frac{\partial}{\partial x} f(t, x)\) reflects the change in the function \(f(t, x)\) due precisely to changes in \(x\). This is critical when your integrand \(f(t, x)\) includes \(x\) explicitly in its formula. In our exercise, the provided integrand, \(\frac{1}{1+t}\), interestingly lacks a direct \(x\) relationship.As a result, the partial derivative with respect to \(x\) becomes zero, simplifying the overall formula significantly. This simplifies the application of Leibniz's rule: we can often ignore the integral of the partial derivative term completely, focusing only on how the limits interact with the integrand.