Problem 28 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$f(x, y)=\int_{y}^{x} \cos \left(t^{2}\right) d t$$

Step-by-Step Solution

Answer

The partial derivatives are \( \frac{\partial f}{\partial x} = \cos(x^2) \) and \( \frac{\partial f}{\partial y} = -\cos(y^2) \).

Step 1: Define the Function

Let the function be defined as follows: \( f(x, y) = \int_{y}^{x} \cos(t^2) \, dt \). Our goal is to compute the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

Step 2: Find Partial Derivative with respect to x

Use the Fundamental Theorem of Calculus, which states that \( \frac{\partial}{\partial x} \left( \int_{a}^{x} g(t) \, dt \right) = g(x) \). Here, \( g(t) = \cos(t^2) \) and the upper limit of integration is \( x \). Thus, \( \frac{\partial f}{\partial x} = \cos(x^2) \).

Step 3: Find Partial Derivative with respect to y

Use the Fundamental Theorem of Calculus again, but applied to the lower limit of integration. If the integral is \( h(y) = \int_{y}^{b} g(t) \, dt \), then \( \frac{dh}{dy} = -g(y) \). Thus, \( \frac{\partial f}{\partial y} = -\cos(y^2) \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is an essential pillar in calculus, linking the concept of differentiation with integration. This theorem comes in two parts.

• The first part states that if a function is continuous over a closed interval \([a, b]\), then the definite integral of the function over that interval is represented by an antiderivative of the function at the bounds of the interval.
• The second part says that if you have an antiderivative \(F(x)\) of a function \(f(x)\) on an interval, then the derivative of \(F\) gives you back the original function \(f\), that is, \(F'(x) = f(x).\)

In simpler terms, the theorem helps us understand how we can "reverse" differentiation through integration. When applying this theorem to find partial derivatives of functions like in our problem, your limits of integration are the key. They determine how the derivative operation is executed.

Integration Concepts

Integration is the process of finding the integral of a function, which can be thought of as the "area under a curve" for a given function. Integration comes in various forms, with the most commonly used being

• definite integration, where you integrate over a specified interval with the result as a number,
• and indefinite integration, which results in a family of functions.

When integrating functions in calculus, particularly those within an integral involving a variable limit as seen in \( f(x, y) = \int_{y}^{x} \cos(t^2) dt \), it's important to accurately handle the limits during differentiation. In the context of partial derivatives, altering the upper or lower limit directly impacts the derivative, requiring the Fundamental Theorem of Calculus for appropriate application. It's crucial to understand these integration mechanisms to correctly solve such problems.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts into multiple dimensions. This expansion introduces new ways of exploring functions that depend on several variables, such as in our exercise where \(f(x, y) = \int_{y}^{x} \cos(t^2) dt\).

Key aspects include:

• Partial derivatives, which measure how a function changes as one variable changes, keeping all other variables constant.
• Gradients, which generalize the concept of derivatives to vector fields, showing the direction of maximum increase of a function.

By focusing on one variable at a time, multivariable calculus helps us simplify and understand complex problems. For instance, when finding the partial derivative with respect to \(x\) and \(y\), we consider how the function changes by altering just one variable while holding the other constant, employing the Fundamental Theorem of Calculus to aid in the differentiation process.