Problem 29 - Calculus: Early Transcendental Functions

Question

find the derivative \(f^{\prime}(x)\) \(f(x)=\int_{0}^{x^{2}}\left(e^{-t^{2}}+1\right) d t\)

Step-by-Step Solution

Answer

The derivative \(f^{\prime}(x)\) is \(2x(e^{-x^{4}} + 1)\).

Step 1: Recognize the Fundamental Theorem of Calculus Part 1

The Fundamental Theorem of Calculus Part 1 states that if a function \(f\) is continuous over the interval \([a, b]\) and \(F\) is an antiderivative of \(f\) on \([a, b]\), then \(\int_{a}^{b}f(x) dx = F(b) - F(a)\). In this case, the function \(f(t) = e^{-t^{2}} + 1\) is continuous over the interval \([0, x^2]\).

Step 2: Use the Chain Rule

Because the upper limit of the integral is the function \(x^2\) rather than simply \(x\), we will need to apply the chain rule. The chain rule states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function.

Step 3: Apply the Fundamental Theorem of Calculus and Chain Rule

Apply the first part of Fundamental Theorem of Calculus, which gives the derivative of the integral as \(f(x^2)\). Then apply the chain rule to take into account the \(x^2\) upper limit. This gives \(f^{\prime}(x) = 2x(f(x^2)) = 2x(e^{-x^{4}} + 1)\).

Key Concepts

Derivative of Integral

The concept of the derivative of an integral is central to understanding the connection between differential and integral calculus. This relationship is elegantly presented in the Fundamental Theorem of Calculus, which is divided into two parts. The first part deals specifically with the derivative of a definite integral. It states that if you have a function that is continuous over an interval and you integrate it over that interval, then the derivative of this integral, with respect to its upper limit is the original function evaluated at that upper limit.

In the context of the given exercise, if we take the integral of the function \( f(t) = e^{-t^{2}} + 1 \) between 0 and \( x^2 \), the derivative of this integral with respect to \( x \) will be \( f(x^2) \), i.e. \( e^{-x^4} + 1 \). This step directly applies the first part of the Fundamental Theorem of Calculus, simplifying the process of finding derivatives of integral expressions and providing a powerful tool for solving a wide range of problems in calculus.

Chain Rule in Calculus

The chain rule is a formula for calculating the derivative of the composition of two or more functions. In other words, if you have a function that is made up of other functions nested inside each other, the chain rule can help you find the derivative of this overall composition. The rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function.

For instance, in our exercise \( f(x) = \int_{0}^{x^{2}} \left(e^{-t^{2}} + 1\right) d t \), the upper limit of the integral is \( x^2 \), which is itself a function of \( x \). To find \( f'(x) \), we have to differentiate \( f(x^2) \) with respect to \( x \) and we do that by using the chain rule. This means taking the derivative of the inner function, \( x^2 \) (which is \( 2x \)), and multiplying it by the derivative of the outer function (which is the result of the first step, \( f(x^2) \)). As a result, we get \( f'(x) = 2x(e^{-x^4} + 1) \).

Antiderivative

An antiderivative, also known as an indefinite integral, of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \). The process of finding an antiderivative is essentially the reverse of differentiation, and there can be many antiderivatives for any given function. However, all antiderivatives of \( f(x) \) differ only by a constant. The importance of antiderivatives in calculus cannot be overstated, as they are essential for solving definite integrals, among other applications.

In our exercise, finding the antiderivative of the function \( f(t) = e^{-t^{2}} + 1 \) would be a step in applying the Fundamental Theorem of Calculus. However, since this function does not have an elementary antiderivative, we address the problem using the theorem directly without explicitly finding the antiderivative.

Continuous Functions

A continuous function is one where small changes in the input produce small changes in the output. There are no sudden jumps or breaks in the graph of the function. In calculus, continuity of a function over an interval is important because it is one of the prerequisites for applying the Fundamental Theorem of Calculus. If a function is continuous on a closed interval \( [a, b] \), it guarantees that the function can be integrated over that interval and that the integral has an antiderivative that is also continuous.

The exercise mentions that the function \( f(t) = e^{-t^{2}} + 1 \) is continuous over the interval \( [0, x^2] \). This piece of information is crucial because it validates the application of the Fundamental Theorem of Calculus. It implies that we can proceed confidently with finding the derivative of the integral, knowing that there are no discontinuities that could otherwise complicate the process.