Problem 24
Question
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2}^{\ln x} e^{-t} d t, x>0 $$
Step-by-Step Solution
Verified Answer
Using Leibniz's Rule, the derivative \( \frac{dy}{dx} = \frac{1}{x^2} \).
1Step 1: Understanding Leibniz's Rule
Leibniz's Rule is used for differentiating an integral whose limits are functions of the variable with respect to which we are differentiating. It states that if \( y(x) = \int_{a(x)}^{b(x)} f(t, x) \, dt \), then \( \frac{dy}{dx} = f(b(x), x) \frac{db}{dx} - f(a(x), x) \frac{da}{dx} + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \).
2Step 2: Identify Components in the Problem
In the given problem, \( a(x) = 2 \), \( b(x) = \ln x \), and the integrand \( f(t) = e^{-t} \) is independent of \( x \). The derivative of the lower limit \( a(x) \) is \( \frac{d}{dx}[2] = 0 \) and the derivative of the upper limit \( b(x) \) is \( \frac{d}{dx}[\ln x] = \frac{1}{x} \). Since \( f(t, x) \) has no \( x \) dependency, \( \frac{\partial}{\partial x} f(t, x) = 0 \).
3Step 3: Apply Leibniz's Rule
Using Leibniz's Rule, we have:\[\frac{dy}{dx} = f(\ln x) \cdot \frac{1}{x} - f(2) \cdot 0 + \int_{2}^{\ln x} 0 \, dt\]Which simplifies to:\[\frac{dy}{dx} = e^{-\ln x} \cdot \frac{1}{x} - 0 + 0\]
4Step 4: Simplify the Expression
To simplify \( e^{-\ln x} \), we use the property that \( e^{-\ln x} = \frac{1}{x} \). Therefore:\[\frac{dy}{dx} = \frac{1}{x} \cdot \frac{1}{x} = \frac{1}{x^2}\]
Key Concepts
Differentiation of IntegralsVariable LimitsCalculus Techniques
Differentiation of Integrals
In calculus, differentiating an integral might sound complex, but Leibniz's rule simplifies this task when faced with integrals that have variable boundaries. Imagine you have an integral where the upper or lower limits are not constants, but functions. This technique allows us to find the derivative of the integral concerning the variable of interest. To put it simply, when you have a function defined as an integral with variable limits, Leibniz's rule helps compute the derivative seamlessly. The rule provides a formula that incorporates the values of the integrand at the boundaries and the derivatives of the limits themselves. This is particularly useful in solving problems where the integral boundaries depend on a variable, often appearing in physics and engineering problems, enhancing our ability to model real-world phenomena.
Variable Limits
Variable limits in integrals can initially appear daunting, especially when trying to find the derivative with respect to these changing boundaries. When limits are functions themselves, the standard rules of integration and differentiation need a bit more specialization in their application, which is where Leibniz's rule comes into play.
- If the upper or lower limit of an integral is a function of the variable of interest, this can affect how the integral behaves as that variable changes.
- In the given exercise, the upper limit is a function of the variable, specifically \(\ln x\), while the lower limit is a constant, \2\.
Calculus Techniques
The exercise of differentiating an integral with variable limits highlights several important calculus techniques. These techniques are crucial for solving complex mathematical problems and understanding phenomena represented in mathematical form. When using Leibniz’s rule, several key aspects of calculus come into play:
- Chain Rule: Useful when taking the derivative of the limits, such as \ln x\.
- Fundamental Theorem of Calculus: Connects differentiation and integration, the two main operations in calculus.
- Property of Exponentials and Logarithms: Simplifying terms like \(e^{-\ln x}\) using identities.
Other exercises in this chapter
Problem 23
If \(\frac{d B}{d t}\) represents the rate of change of biomass at time \(t\), explain what $$ \int_{1}^{6} \frac{d B}{d t} d t $$ means.
View solution Problem 23
Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$
View solution Problem 24
Let \(N(t)\) denote the size of a population at time \(t\), and assume that $$ \frac{d N}{d t}=f(t) $$ Express the cumulative change of the population size in t
View solution Problem 24
Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$
View solution