Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all first partial derivatives of each function. \(f(x, y)=e^{x} \cos y\)

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = e^x \cos y \) and \( \frac{\partial f}{\partial y} = -e^x \sin y \).

Step 1: Identify the function

The given function is \( f(x, y) = e^x \cos y \). We need to find the first partial derivatives with respect to \( x \) and \( y \).

Step 2: Partial Derivative with Respect to \( x \)

To find the partial derivative of \( f(x, y) \) with respect to \( x \), treat \( y \) as a constant. The derivative of \( e^x \) with respect to \( x \) is \( e^x \), and the cosine part remains as it is. Thus, \( \frac{\partial f}{\partial x} = e^x \cos y \).

Step 3: Partial Derivative with Respect to \( y \)

Now find the partial derivative of \( f(x, y) \) with respect to \( y \), treating \( x \) as a constant. The derivative of \( \cos y \) with respect to \( y \) is \( -\sin y \). Therefore, \( \frac{\partial f}{\partial y} = e^x (-\sin y) = -e^x \sin y \).

Step 4: Conclusion

We have found both partial derivatives of the function. \( \frac{\partial f}{\partial x} = e^x \cos y \) and \( \frac{\partial f}{\partial y} = -e^x \sin y \).

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics that studies rates of change and how quantities vary with respect to others. It is a foundational subject that enables us to solve complex problems across various scientific disciplines. There are two main branches of calculus: differential calculus and integral calculus.

• Differential Calculus: Focuses on the concept of the derivative, which measures how a function changes as its input changes. It helps understand slopes of curves and rates of change.
• Integral Calculus: Deals with accumulation of quantities, such as areas under a curve and total accumulated change.

In calculus, particularly in the context of the problem we're exploring, understanding how functions change when altering their variables is essential for finding derivatives, which lead to solutions of real-world problems.

Exploring Multivariable Functions

Multivariable functions involve more than one variable, creating surfaces instead of simple lines or curves. For instance, the function given in the exercise, \( f(x, y) = e^x \cos y \), involves the variables \( x \) and \( y \). Changes in these variables can result in different values of the function, much like how altering ingredients in a recipe can change the final dish.

• These functions map points from a plane (\( x, y \)) to a certain value, similar to plotting different altitudes on a topographic map.
• Understanding multivariable functions is crucial in fields such as physics, engineering, and economics, where systems with multiple influencing factors are common.

Using calculus to differentiate these functions, as in finding partial derivatives, provides valuable insights into how each variable uniquely affects the function.

Differentiation and Partial Derivatives

Differentiation is the process of finding the derivative of a function, which provides information about its rate of change. In multivariable calculus, this leads to the concept of partial derivatives. These derivatives measure how a multivariable function changes with respect to one variable, while keeping others constant.

• Example: For \( f(x, y) = e^x \cos y \), when finding the partial derivative with respect to \( x \), treat \( y \) as a constant, resulting in \( \frac{\partial f}{\partial x} = e^x \cos y \).
• Conversely, for partial differentiation with respect to \( y \), \( x \) is constant, giving \( \frac{\partial f}{\partial y} = -e^x \sin y \).

This method is essential for optimizing functions and modeling in multiple dimensions, allowing for greater precision in predicting changes and responses in systems.

Trigonometric Functions in Calculus

Trigonometric functions like \( \cos \) and \( \sin \) are essential in calculus, often arising in problems involving oscillations, waves, and circular movements. In the given function \( f(x, y) = e^x \cos y \), \( \cos y \) represents how the function value varies with \( y \), influenced by changes in angle or phase.

• Understanding how to differentiate these functions is key, as seen in the exercise where \( \cos y \) becomes \( -\sin y \) when differentiated with respect to \( y \).
• These derivatives are pivotal in accurately modeling behaviors in physics and engineering, where wave patterns and periodic motions need precise calculations.

Mastering trigonometric functions' behavior in calculus contexts enables solutions to more complex real-world problems, enhancing fields reliant on cyclic or oscillatory processes.