Problem 458 - Calculus Volume 1

Question

Find the derivative. $$\frac{d}{d x} \int_{0}^{\cos x} e^{t^{2}} d t$$

Step-by-Step Solution

Answer

The derivative is \(-\sin x \cdot e^{(\cos x)^2}\).

Step 1: Recognize the Exercise Type

The problem requires finding the derivative of a function that includes an integral with a variable limit. This is a classic example involving the Fundamental Theorem of Calculus and Leibniz's Rule.

Step 2: Apply the Fundamental Theorem of Calculus

First, understand that if we have a function of the form \( F(x) = \int_{a(x)}^{b(x)} f(t) \, dt \), then its derivative, \( \frac{d}{dx} F(x) \), can be found using Leibniz's rule.

Step 3: Identify Function Components

In this exercise, our function is \( G(x) = \int_0^{\cos x} e^{t^2} \, dt \). Here, \( f(t) = e^{t^2} \), the lower limit is a constant \( a(x) = 0 \), and the upper limit is a function \( b(x) = \cos x \).

Step 4: Use Leibniz's Rule

Leibniz's Rule gives us the formula for differentiation: \[ \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]. In our example, \( a(x) = 0 \) gives \( a'(x) = 0 \), so the second term disappears.

Step 5: Differentiate with Respect to x

For \( G(x) = \int_0^{\cos x} e^{t^2} \, dt \), apply Leibniz's Rule: \( \frac{d}{dx} G(x) = e^{(\cos x)^2} \cdot \frac{d}{dx} (\cos x) \). Calculate the derivative of \( \cos x \), which is \( -\sin x \).

Step 6: Calculate the Final Result

Substitute \( b'(x) \) back into our expression. The derivative is \( e^{(\cos x)^2} \cdot (-\sin x) = -\sin x \cdot e^{(\cos x)^2} \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus, linking the concept of differentiation and integration. It comes in two parts, with the first part stating that if a function is continuous over a closed interval \([a, b]\), then the integral of the function over that interval is tied to its antiderivative. In simple terms, it means that the process of integrating and then differentiating
will return the function to its original form. This theorem essentially states that differentiation and integration are inverse operations.

• Part 1: If \(F(x)\) is an antiderivative of \(f(x)\) on an interval \(I\), then \(F(x)\) is given by \(F(x) = \int_{a}^{x} f(t) \, dt\) for all \(x\) in \(I\).
• Part 2: If \(f\) is continuous on \([a, b]\) and \(F\) is any antiderivative of \(f\) on \([a, b]\), then \(\int_{a}^{b} f(t) \, dt = F(b) - F(a)\).

When problems involve variable limits of integration, like our exercise with the integral \(\int_{0}^{\cos x} e^{t^2} \, dt\), we apply this theorem to find the derivative of an integral, revealing the dynamic connection between integration and differentiation.

Leibniz's Rule

Leibniz's Rule is a powerful tool used when differentiating integrals with variable limits. This rule extends the Fundamental Theorem of Calculus to cases where the limits of integration are not constant, but functions of the variable of differentiation.
For a function given by \(F(x) = \int_{a(x)}^{b(x)} f(t) \, dt\), Leibniz's Rule provides a method to find its derivative:
\[\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\]This formula reflects the contribution to the derivative from both the upper and lower limits of integration becoming dynamic.

In simpler words:

• The term \(f(b(x)) \cdot b'(x)\) accounts for the change in the upper limit owing to the derivative \(b'(x)\).
• Since in our exercise \(a(x) = 0\) and its derivative \(a'(x) = 0\), the corresponding term disappears, simplifying the differentiation of such integrals.

This rule is instrumental in solving the given problem where the upper limit is a variable \(\cos x\), making it crucial to employ Leibniz's Rule to find the correct derivative.

Derivative of Integral

When faced with differentiating an integral that has a variable limit, the concept of taking the Derivative of an Integral becomes essential. This involves combining the Fundamental Theorem of Calculus and Leibniz's Rule.
In our problem, we use this technique to find the derivative of \(G(x) = \int_{0}^{\cos x} e^{t^2} \, dt\).

• The key is finding \(G'(x)\) by applying Leibniz's Rule which guides us to use the function inside the integral at the variable upper limit.
• Recognize that \(G(x)\) is dependent on the upper limit \(b(x) = \cos x\).

The steps are as follows:
1. Identify \(f(t) = e^{t^2}\), which is integrated from \(0\) to \(\cos x\).
2. Calculate the derivative of the upper limit, \(b'(x) = \frac{d}{dx}(\cos x) = -\sin x\).
3. Substitute \(b'(x)\) and \(f(b(x)) = e^{(\cos x)^2}\) into the formula:
\[G'(x) = e^{(\cos x)^2} \cdot (-\sin x) = -\sin x \cdot e^{(\cos x)^2}\]This process reveals that even though the function is wrapped around an integral, differentiation can still be performed, revealing the intricate connection between the original function and its derivative.