Problem 35 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 3x^2 \ln(x^3 - 3) - 2x \ln(x^2 - 3) \)

Step 1: Identify the problem structure

We are given a definite integral with variable limits. The task is to differentiate this function with respect to the variable \( x \), using Leibniz's rule for differentiating under the integral sign.

Step 2: Recall Leibniz's Rule

Leibniz's Rule for differentiating under the integral sign states: \[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) \, dt \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \]Here, \( f(t) = \ln(t-3) \), and there is no explicit dependence on \( x \) in the integrand, so \( \frac{\partial}{\partial x} f(t) = 0 \).

Step 3: Evaluate \(f(b(x))\) and \(f(a(x))\)

Substitute \( b(x) = x^3 \) and \( a(x) = x^2 \) into the integrand function \( f(t) = \ln(t-3) \):1. \( f(b(x)) = \ln(x^3 - 3) \)2. \( f(a(x)) = \ln(x^2 - 3) \)

Step 4: Differentiate the limits

Differentiate the upper limit \( b(x) = x^3 \) and the lower limit \( a(x) = x^2 \) with respect to \( x \):1. \( b'(x) = 3x^2 \)2. \( a'(x) = 2x \)

Step 5: Apply Leibniz's Rule and solve

Substitute the expressions found into Leibniz's Rule:\[ \frac{dy}{dx} = \ln(x^3 - 3) \cdot 3x^2 - \ln(x^2 - 3) \cdot 2x \]There is no \( \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t) \, dt \) term, as \( \frac{\partial}{\partial x} f(t) = 0 \).

Step 6: Simplify the expression

The derivative is \[ \frac{dy}{dx} = 3x^2 \ln(x^3 - 3) - 2x \ln(x^2 - 3) \]This is the simplified form of the differentiated function using Leibniz's rule.

Key Concepts

Differentiating Under the Integral Sign

When faced with a problem that involves a definite integral whose limits vary with respect to a variable, we can simplify our work using Leibniz's Rule. This powerful technique falls under the umbrella of differentiating under the integral sign. Essentially, Leibniz's Rule provides a method to differentiate an integral whose upper and lower limits are functions of a single variable, often specified as \(x\). The rule can be formulated as follows:\[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) \, dt \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \]Here's a breakdown of the needed terms in this formula:- **\(f(b(x))\)** and **\(f(a(x))\):** are the function \(f(t)\) computed at the upper and lower limits, respectively.- **\(b'(x)\)** and **\(a'(x)\):** are the derivatives of the functions defining the upper and lower limits with respect to the variable \(x\).- The term involving the integral of \(\frac{\partial}{\partial x} f(t, x)\) represents the change in the function under the integral with respect to \(x\), but this is usually zero if \(f(t)\) does not explicitly depend on \(x\).In our exercise, where \(y = \int_{x^{2}}^{x^{3}} \ln(t-3) \, dt\), the integrand \(ln(t-3)\) does not include \(x\), making the last integral term vanish, simplifying our calculations.

Variable Limits of Integration

In the context of Leibniz's Rule, variable limits of integration are the endpoints of our integral, expressed as functions that depend on our variable of differentiation. In the given exercise, these limits are expressed as \(a(x) = x^2\) and \(b(x) = x^3\). This dependency allows us to see how changes in \(x\) affect the boundaries of integration.Consider the mechanics of computing these derivatives:- **Compute \(b'(x)\):** This represents the derivative of the upper limit, \(x^3\), with respect to \(x\). By differentiation, we find \(b'(x) = 3x^2\).- **Compute \(a'(x)\):** Similarly, the derivative of the lower limit, \(x^2\), with respect to \(x\) is \(a'(x) = 2x\).These derivatives are crucial since they directly influence the final derivative of the integral. They help account for the change in the area under the curve as the limits move. With these derivatives, we can accurately apply Leibniz's Rule to find how the definite integral changes as \(x\) changes.

Definite Integral Differentiation

Definite integral differentiation involves finding the derivative of an integral where the limits are not just constants, but functions of a variable. In problems where the limits vary, as in the given exercise, Leibniz's Rule serves as a robust tool to handle differentiation efficiently.Here's the stepwise breakdown of the process:- **Evaluate \(f(b(x))\) and \(f(a(x))\):** By substituting \(b(x) = x^3\) and \(a(x) = x^2\) back into the integrand \(\ln(t-3)\), we have: - \(f(x^3) = \ln(x^3 - 3)\) - \(f(x^2) = \ln(x^2 - 3)\)- **Apply Leibniz's Rule:** Gather all elements: - Multiply each function at the limit by the derivative of the limit: - \(3x^2 \cdot \ln(x^3 - 3)\) - \(-2x \cdot \ln(x^2 - 3)\) (note the negative sign since it is the lower limit)- **Combine terms:** The result becomes the derivative of the original integral:\[ \frac{dy}{dx} = 3x^2 \ln(x^3 - 3) - 2x \ln(x^2 - 3) \]This expression represents how the integral of \(\ln(t-3)\) from \(x^2\) to \(x^3\) changes with respect to \(x\). Understanding this concept allows for accurate computation in mathematical problems involving both limits and integrand changes.