Question

Let \(f(x, y)=3 x^{2} y e^{x-y}\) and \(g(x, y)=\) \(3 x y^{2} e^{y-x}\). Find each of the following. $$ g_{x}(x, y) $$

Short answer

The result is \(\frac{∂g}{∂x} = 3y^{2}e^{y-x}(1-x)\).

Step-by-step solution

Identify the Functions

The function \(g(x, y)\) is indeed a product of functions. We can denote these as \(u(x,y) = 3xy^{2}\) and \(v(x, y) = e^{y-x}\).

Compute the Derivatives

Next, we have to compute the partial derivatives \(\frac{∂u}{∂x}\) and \(\frac{∂v}{∂x}\). For the first function, \(\frac{∂u}{∂x} = 3y^{2}\). For the second function, we use the chain rule. First, the derivative of \(e^{x}\) is simply \(e^{x}\), but since we have \(e^{y-x}\), the derivative of the exponential gets multiplied by the derivative of \(y-x\) (the function inside the exponent). This gives us \(\frac{∂v}{∂x} = -e^{y-x}\).

Apply the Product Rule

Since the function \(g(x, y)\) is a product of functions, the product rule is used to get its derivative. The product rule in this context is given by \(\frac{∂g}{∂x} = u\frac{∂v}{∂x} + v\frac{∂u}{∂x}\). Substituting, we get: \(\frac{∂g}{∂x} = 3xy^{2}(-e^{y-x}) + e^{y-x}(3y^{2}) = -3xy^{2}e^{y-x} + 3y^{2}e^{y-x}\).

Simplify the Result

The resulting expression could be further simplified by factoring out common terms. We can factor out \(3y^{2}e^{y-x}\) to simplify as, \(\frac{∂g}{∂x} = 3y^{2}e^{y-x}(-x+1) = 3y^{2}e^{y-x}(1-x)\).

Key concepts

Product Rule in Partial Derivatives

The product rule is a fundamental rule in calculus that is also applicable when taking the partial derivatives of a function presented as a product of two distinct sub-functions. Just as in single-variable calculus, the rule states that the derivative of a product of two functions is not simply the product of their derivatives. Instead, it requires an additional step of calculating "mixed" derivatives between the functions.

In our example, the function given is a product of two functions: - \( u(x,y) = 3xy^{2} \) - \( v(x,y) = e^{y-x} \)

When differentiating with respect to a particular variable, such as \( x \), the product rule is employed: \( \frac{\partial (uv)}{\partial x} = u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x} \). Using the above example, the differentiated expression becomes \(-3xy^{2}e^{y-x} + 3y^{2}e^{y-x} \).

This approach allows for capturing the interaction between the changes in both sub-functions with regard to the variable \( x \). The method brings balance by combining the rates of change from each function separately as they contribute to the product.

Using the Chain Rule in Multivariable Functions

The chain rule is a tool used in calculus to differentiate composite functions, especially when a function resides inside another function. This rule is crucial for finding derivatives of complex functions, and in multivariable calculus, it extends to functions with more than one variable. When dealing with partial derivatives, it means taking derivatives of functions inside another function with respect to each variable and considering how each piece relates through multiplication.

In our worked example, the chain rule is applied to the exponential function \( v(x, y) = e^{y-x} \). - The derivative of \( e^{y-x} \) comes from differentiating the inner function \( y - x \), yielding - The inner derivative results in \( -1 \) because of the simple subtraction expression \( y-x \). - Consequently, the partial derivative \( \frac{\partial v}{\partial x} = -e^{y-x} \)

The process highlights the necessity of performing an extra calculation for any function nested within another. This ensures that the changes in the inner function’s variable contribution to the outer functional response are accurately captured.

Understanding Multivariable Calculus

Multivariable calculus extends the principles and techniques from single-variable calculus to higher dimensions, typically involving functions with more than one variable, such as \( f(x, y) \), \( g(x, y) \), etc. In such cases, rather than a simple derivative, we examine partial derivatives, which measure how a function changes as only one of the variables is changed, holding the others constant.

In the context of our solution, the function \( g(x, y) = 3xy^{2} e^{y-x} \) requires differentiation with respect to \( x \). This leads us to compute the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial v}{\partial x} \) separately. These derivatives capture the extent to which the respective portions of the function change as \( x \) varies.

Ultimately, knowledge of multivariable calculus is essential for understanding more complex systems modeled by changes in many directions, and it's heavily used in fields such as physics, engineering, and computer science. The tools from single-variable calculus, namely the product rule and chain rule, are adapted to fit these higher-dimensional scenarios, integrating well within the framework of partial derivatives.