Problem 122 - Calculus Volume 3

Question

Find \(f_{y}(x, y)\) for \(f(x, y)=e^{x y} \cos (x) \sin (y)\).

Step-by-Step Solution

Answer

The partial derivative is \( f_{y}(x, y) = xe^{xy} \cos(x) \sin(y) + e^{xy} \cos(x) \cos(y) \).

Step 1: Identify the Function and Target

We are given the function \(f(x, y) = e^{xy} \cos(x) \sin(y)\) and we need to find the partial derivative with respect to \(y\), which is \(f_y(x, y)\).

Step 2: Apply the Product Rule

Notice that \(f(x, y)\) is a product of three functions: \(e^{xy}\), \(\cos(x)\), and \(\sin(y)\). The partial derivative \(f_y(x, y)\) requires the use of the product rule. We treat \(\cos(x)\) as a constant with respect to \(y\).

Step 3: Differentiate the Exponential Component

Differentiate \(e^{xy}\) with respect to \(y\). Using the chain rule, the derivative is \(xe^{xy}\).

Step 4: Differentiate the Sine Function

The derivative of \(\sin(y)\) with respect to \(y\) is \(\cos(y)\).

Step 5: Combine Results Using the Product Rule

Using the product rule for differentiation, we have the following: \[ f_y(x, y) = \cos(x) \left( xe^{xy} \sin(y) \right) + e^{xy} \cos(x) \cos(y). \]

Step 6: Simplify the Expression

The expression simplifies to:\[ f_y(x, y) = xe^{xy} \cos(x) \sin(y) + e^{xy} \cos(x) \cos(y). \]

Key Concepts

Product Rule

The product rule is a fundamental principle in calculus, especially when dealing with differentiation. It is used when you need to differentiate a function which is the product of two or more functions. The product rule states that the derivative of a product of two functions, say, \( u(x) \) and \( v(x) \), is given by:

• \( (uv)' = u'v + uv' \)

In our exercise, we apply this rule to the function \( f(x, y) = e^{xy} \cos(x) \sin(y) \), which is a product of three sub-functions. When considering the partial derivative with respect to \( y \), treat terms involving \( x \) without \( y \) as constants. We thus apply the product rule sequentially to each term, focusing on the partial differentiation of elements involving \( y \). This application allows us to break down the problem systematically to avoid missing any components associated with \( y \).

Chain Rule

The chain rule is another critical concept when dealing with derivatives, mainly when functions are composed within each other. It is a way to differentiate compositions of functions. If you have a composite function \( g(f(x)) \), the chain rule tells us that the derivative is:

• \( (g(f(x)))' = g'(f(x)) \cdot f'(x) \)

In the context of the given exercise, we use the chain rule when differentiating \( e^{xy} \) with respect to \( y \). Here, consider \( xy \) as an inner function. Applying the chain rule results in multiplying the derivative of the outer function \( e^{something} \) by the derivative of the inside function \( xy \), that is, \( x \cdot e^{xy} \). This application is crucial for obtaining the correct derivative for terms involving exponential functions which depend on a product of variables.

Exponential Functions

Exponential functions are a special type of function characterized by their constant growth rate. These functions typically involve the constant \( e \) (approximately 2.718). A standard exponential function is \( e^x \). The derivative of any exponential function \( e^{something} \) is unique because it mirrors the original function's structure:

• \( \frac{d}{dx} e^x = e^x \)

In our exercise, we deal with the term \( e^{xy} \). Differentiating \( e^{xy} \) requires using the chain rule since \( xy \) depends on the variable \( y \). Thus, the partial derivative is computed as \( xe^{xy} \) due to the presence of \( x \) as a multiplying factor from the derivative of the exponent \( xy \). This operation demonstrates the powerful nature of exponential functions in derivatives simplifying into exponential forms, maintaining their fundamental characteristics.

Trigonometric Functions

Trigonometric functions such as sine and cosine are periodic and frequently used ratios derived from the sides of a right triangle. Differentiation of these functions is fundamental in calculus due to their cyclic nature. The derivatives of the sine and cosine functions are as follows:

• \( \frac{d}{dy} \sin(y) = \cos(y) \)
• \( \frac{d}{dy} \cos(y) = -\sin(y) \)

In the exercise, we need to compute the partial derivative of \( \sin(y) \) with respect to \( y \), resulting in \( \cos(y) \). Subsequently, we multiply this result by other components in the expression as dictated by the product rule. The involvement of trigonometric functions in differentiation illustrates their inherent properties and showcases the vital role they play in forming more complex functions involving variable-dependent derivatives.