Problem 56 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\int_{-1}^{x^{1 / 2}} \sin ^{-1} t d t$$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{2} x^{-1/2} \sin^{-1}(x^{1/2}) \).

Step 1: Identifying the Type of Problem

We have been asked to differentiate a function defined by an integral with upper limit dependent on x, specifically in the form \( y = \int_{-1}^{x^{1/2}} \sin^{-1} t \, dt \). This involves applying the Fundamental Theorem of Calculus combined with the Chain Rule as \( y \) is a function of \( x \) through the upper limit.

Step 2: Applying the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(t) = \int_{a}^{t} f(u) \, du \), then \( \frac{d}{dt} F(t) = f(t) \). Therefore, for \( y = \int_{-1}^{x^{1/2}} \sin^{-1} t \, dt \), the derivative \( \frac{d}{dx^{1/2}} \) is \( \sin^{-1}(x^{1/2}) \).

Step 3: Applying the Chain Rule

Let \( u = x^{1/2} \) so \( du/dx = \frac{1}{2} x^{-1/2} \). Applying the Chain Rule gives \( \frac{d}{dx} y = \sin^{-1}(x^{1/2}) \cdot \frac{du}{dx} = \sin^{-1}(x^{1/2}) \cdot \frac{1}{2} x^{-1/2} \).

Step 4: Writing the Final Expression

After simplifying, we find \( \frac{dy}{dx} = \frac{1}{2} x^{-1/2} \sin^{-1}(x^{1/2}) \). This expression represents the rate of change of \( y \) with respect to \( x \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in the field of calculus, providing a connection between differentiation and integration. It consists of two main parts:

• The first part states that if a function is continuous over an interval, then its indefinite integral can be "reversed" by differentiation.
• The second part asserts that if we have a function defined as an integral with a variable upper limit, the derivative of this integral with respect to the upper limit is simply the integrand evaluated at this upper limit.

In the original exercise, the function \( y \) is defined as an integral with a variable in the upper limit. Specifically, it is given as \( y = \int_{-1}^{x^{1/2}} \sin^{-1} t \, dt \). Applying the second part of the Fundamental Theorem, we derive that the rate of change of \( y \) with respect to the upper limit \( x^{1/2} \) is \( \sin^{-1}(x^{1/2}) \). This result is at the heart of solving the problem, showing the power of this theorem to simplify the differentiation of integrals.

Chain Rule

The Chain Rule is a fundamental tool in calculus used to calculate the derivative of composite functions. When a function has an inner function nested inside another function, the Chain Rule allows us to determine the overall rate of change.

• Let's say we have a function \( y = f(g(x)) \), where \( g(x) \) is the inner function.
• According to the Chain Rule, the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of \( f \) with respect to \( g \) by the derivative of \( g \) with respect to \( x \): \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In our exercise, the necessity of the Chain Rule arises because the upper limit of the integral, \( x^{1/2} \), is itself a function of \( x \). Defining \( u = x^{1/2} \), we can differentiate it as \( \frac{du}{dx} = \frac{1}{2} x^{-1/2} \). By applying the Chain Rule, we relate this result with the derivative found from the Fundamental Theorem: \( \frac{d}{dx} y = \sin^{-1}(x^{1/2}) \cdot \frac{1}{2} x^{-1/2} \). This combination offers a complete solution to find \( \frac{dy}{dx} \).

Integration with Variable Limits

When dealing with integrals, variable limits add an extra layer of complexity, yet they frequently appear in calculus problems. Here, the integral defines a function with one of its limits as a variable function of \( x \).

• In simple terms, this means the bounds of integration are not fixed numbers, but rather expressions involving the variable whose change we’re tracking.
• Such cases require careful handling since the variable limits affect the differentiation of the integral.

In the given exercise, the integration ranges from \( -1 \) to \( x^{1/2} \). The presence of \( x^{1/2} \) indicates that the upper limit changes as \( x \) changes. To differentiate this type of integral, the Fundamental Theorem of Calculus must be combined with the Chain Rule. By first computing the derivative as if \( x^{1/2} \) were a constant and then adjusting for its variable nature using the Chain Rule, we expertly account for how changes in \( x \) influence the integral's value. This leads to the final differentiated expression \( \frac{1}{2} x^{-1/2} \sin^{-1}(x^{1/2}) \), a textbook application of these concepts.