Problem 23 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ f(x, y)=x e^{x^{3} y} $$

Step-by-Step Solution

Answer

The first partial derivatives are \( f_x(x, y) = e^{x^3 y} (1 + 3x^3 y) \) and \( f_y(x, y) = x^4 e^{x^3 y} \).

Step 1: Understanding Partial Derivatives

Partial derivatives involve taking the derivative of a function with respect to one variable while keeping other variables constant. Here, we will find the partial derivatives of the function \( f(x, y) = x e^{x^3 y} \) 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 \( x \), we treat \( y \) as a constant. The product rule is needed since our function is a product of \( x \) and \( e^{x^3 y} \). The derivative of \( x \) is 1, and the derivative of \( e^{x^3 y} \) with respect to \( x \) is \( 3x^2 y e^{x^3 y} \). Thus, the partial derivative is:\[ f_x(x, y) = 1 imes e^{x^3 y} + x imes 3x^2 y e^{x^3 y} = e^{x^3 y} + 3x^3 y e^{x^3 y} \]Simplify to:\[ f_x(x, y) = e^{x^3 y} (1 + 3x^3 y) \]

Step 3: Partial Derivative with respect to y

To find the partial derivative of \( f(x, y) \) with respect to \( y \), we treat \( x \) as a constant. The derivative of \( e^{x^3 y} \) with respect to \( y \) is \( x^3 e^{x^3 y} \). Thus, the partial derivative is:\[ f_y(x, y) = x imes x^3 e^{x^3 y} = x^4 e^{x^3 y} \]

Key Concepts

Product Rule

The product rule is essential when we need to differentiate expressions where two or more functions are multiplied together. In essence, it states that if you have a function that is the product of two other functions, then the derivative is given by:

• The derivative of the first function times the second function
• Plus the first function times the derivative of the second function

This is written mathematically as: \[ \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \]where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives.
This rule is particularly useful in multivariable calculus when we are dealing with partial derivatives. It helps simplify complex expressions into more manageable parts. For example, in the function \(f(x, y) = x e^{x^3 y} \), applying the product rule allows us to differentiate with respect to one variable while keeping the other constant. This is crucial for accurately calculating partial derivatives without losing components of the expression.

Exponential Function

Exponential functions, represented as \(e^{x}\), are functions where the independent variable is the exponent of a constant base \( e \), which is approximately equal to 2.71828. These functions have unique properties:

• The rate of growth of an exponential function is proportional to its current value.
• The derivative of \(e^{x}\) with respect to \(x\) is itself: \(\frac{d}{dx}(e^{x}) = e^{x}\).
• When the exponent is more complex, such as \(e^{x^3 y}\), differentiation involves the chain rule.

For instance, the calculation of \(\frac{\partial}{\partial x} e^{x^3 y}\) requires taking the derivative of the exponent \(x^3 y\) and multiplying it by the original exponential term, an application of the chain rule within the context of exponential functions. Exponential functions are ubiquitous in calculus due to their detailed structure and application in modeling growth processes or decay in real-life phenomena. Understanding these well prepares you for more challenging mathematical scenarios involving exponential terms.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions with multiple variables. This area of mathematics becomes crucial in exploring scenarios where outcomes depend on several factors at once.
In multivariable calculus, functions take the form \(f(x, y, z, \ldots)\), allowing us to compute partial derivatives. Each partial derivative represents how the function changes as we vary one of those variables while keeping the others fixed.

• Partial derivatives are calculated similarly to regular derivatives but focus on one variable at a time.
• This is useful in many fields such as physics and engineering, where systems often depend on multiple inputs.

For example, `f(x, y) = x e^{x^3 y}` requires finding \(\partial f/\partial x\) and \(\partial f/\partial y\), each showing how the function behavior shifts with changes in \(x\) or \(y\) respectively, assuming the other remains constant. These concepts are foundational for more advanced topics like gradient vectors and optimization in higher dimensions.