Problem 75
Question
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.\(h(x)=\int_{2}^{1 / x} \arctan t d t\)
Step-by-Step Solution
Verified Answer
\(-\frac{\arctan\left(\frac{1}{x}\right)}{x^2}\)
1Step 1: Identify the Fundamental Theorem of Calculus Part 1
The first part of the Fundamental Theorem of Calculus states that if \( G(x) = \int_{a}^{x} f(t) \, dt \), then \( \frac{d}{dx} G(x) = f(x) \). This applies when the upper limit of integration is \( x \). In our problem, the upper limit is \( \frac{1}{x} \), meaning an adjustment is needed.
2Step 2: Apply a Change of Variables
We have \( h(x)=\int_{2}^{\frac{1}{x}} \arctan t \, dt \). The upper limit involves the variable \( x \). Let's set it as a function \( u(x) = \frac{1}{x} \), so the problem involves a change of variables.
3Step 3: Derivative of the Integral with a Variable Limit
According to the chain rule, if \( u(x) = \frac{1}{x} \), then \( \frac{d}{dx} h(x) = \arctan(u(x)) \cdot u'(x) \), where \( u'(x) = -\frac{1}{x^2} \). Integrating this yields \( \frac{d}{dx} h(x) = \arctan\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \).
4Step 4: Simplify and Finalize Derivative
Finally, perform the multiplication to simplify the expression. The derivative becomes \( -\frac{\arctan\left(\frac{1}{x}\right)}{x^2} \).
Key Concepts
IntegrationDerivativeChain RuleChange of Variables
Integration
Integration is the process of finding the integral of a function. It is essentially the opposite operation of differentiation. In this context, integration allows us to determine the cumulative sum or area under the curve of a function. When given an integral such as \(\int_{a}^{b} f(t) \, dt\), we use integration to find the total area between the curve \(f(t)\), the x-axis, and the vertical lines \(t = a\) and \(t = b\).
In our specific problem, the function \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\) involves integrating the function \(\arctan t\) between the limits \(t = 2\) and \(t = \frac{1}{x}\).
This setup is a classic example where the Fundamental Theorem of Calculus Part 1 is useful, as it relates integration and differentiation directly.
In our specific problem, the function \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\) involves integrating the function \(\arctan t\) between the limits \(t = 2\) and \(t = \frac{1}{x}\).
This setup is a classic example where the Fundamental Theorem of Calculus Part 1 is useful, as it relates integration and differentiation directly.
Derivative
Derivatives represent the rate at which a function is changing at any given point. Differentiation is the process of calculating a derivative. It is a cornerstone concept in calculus, often used to find how a quantity changes with respect to another variable.
In the context of our problem, we need to find the derivative \(\frac{d}{dx}h(x)\) of the integral \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\). This derivative tells us how the value of this integral changes as the upper limit, \(\frac{1}{x}\), changes.
The main challenge here is dealing with the variable upper limit of integration, which requires additional technique beyond basic differentiation.
In the context of our problem, we need to find the derivative \(\frac{d}{dx}h(x)\) of the integral \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\). This derivative tells us how the value of this integral changes as the upper limit, \(\frac{1}{x}\), changes.
The main challenge here is dealing with the variable upper limit of integration, which requires additional technique beyond basic differentiation.
Chain Rule
The chain rule is a powerful tool in differentiation used when dealing with composite functions—functions within functions. It allows us to differentiate a function based on the derivative of its inner and outer functions.
In our solution, the chain rule becomes particularly important. Since the upper limit of our integral \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\) is a function of \(x\), denoted as \(u(x) = \frac{1}{x}\), the chain rule helps us determine how to properly differentiate.
With the chain rule, we find that \(\frac{d}{dx}h(x) = \arctan\left(\frac{1}{x}\right) \cdot u'(x)\). The derivative of \(u(x)\) is \(u'(x) = -\frac{1}{x^2}\), leading to the complete derivative expression \(-\frac{\arctan\left(\frac{1}{x}\right)}{x^2}\).
This method allows us to successfully differentiate even with changing limits.
In our solution, the chain rule becomes particularly important. Since the upper limit of our integral \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\) is a function of \(x\), denoted as \(u(x) = \frac{1}{x}\), the chain rule helps us determine how to properly differentiate.
With the chain rule, we find that \(\frac{d}{dx}h(x) = \arctan\left(\frac{1}{x}\right) \cdot u'(x)\). The derivative of \(u(x)\) is \(u'(x) = -\frac{1}{x^2}\), leading to the complete derivative expression \(-\frac{\arctan\left(\frac{1}{x}\right)}{x^2}\).
This method allows us to successfully differentiate even with changing limits.
Change of Variables
Changing variables is a technique used in both integration and differentiation to simplify a problem by substituting variables. This tactic often makes complex problems more manageable by using new variables that fit the situation better.
For the function \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\), a change of variables is necessary due to the variable upper limit of integration. We define \(u(x) = \frac{1}{x}\) to create a standard form that we can differentiate more easily.
By substituting \(u(x)\) for \(\frac{1}{x}\), we transform the problem into one where the Fundamental Theorem of Calculus and the chain rule can be applied effectively. This change not only simplifies the differentiation process but also aligns the integral within a usable method.
For the function \(h(x) = \int_{2}^{\frac{1}{x}} \arctan t \, dt\), a change of variables is necessary due to the variable upper limit of integration. We define \(u(x) = \frac{1}{x}\) to create a standard form that we can differentiate more easily.
By substituting \(u(x)\) for \(\frac{1}{x}\), we transform the problem into one where the Fundamental Theorem of Calculus and the chain rule can be applied effectively. This change not only simplifies the differentiation process but also aligns the integral within a usable method.
- Identify the function to change, such as \(u(x) = \frac{1}{x}\).
- Substitute \(u\) into the function and proceed with solving.
Other exercises in this chapter
Problem 72
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \(g(r)=\int_{0}^{r} \sqrt{x^{2}+4} d x\)
View solution Problem 74
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.\(G(x)=\int_{x}^{1} \cos \sqrt{t} d t\)
View solution Problem 76
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \(h(x)=\int_{0}^{x^{2}} \sqrt{1+r^{3}} d r\)
View solution Problem 77
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \(y=\int_{0}^{\tan x} \sqrt{t+\sqrt{t}} d t\)
View solution