Problem 17 - Calculus an Applied Approach

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. $$ g_{x}(x, y) $$

Step-by-Step Solution

Answer

\(\partial g/\partial x = 3y^2e^{y-x}(1 - x)\)

Step 1: Identify the sub-functions

The function \( g(x,y) = 3xy^2e^{y-x} \) is composed of three sub-functions: \( u(x,y) = 3x \), \( v(x,y) = y^2 \), and \( w(x,y) = e^{y-x} \)

Step 2: Differentiate the sub-functions with respect to x

Differentiating the first sub-function \(u(x,y)\) with respect to 'x' while treating 'y' as constant gives \(\partial u/\partial x = 3\). \nThe second sub-function \(v(x,y)\) does not depend on 'x', so its derivative is zero. That is, \(\partial v/\partial x = 0\). \nThe third sub-function \(w(x,y) = e^{y-x}\) involves the variable 'x' in its exponent. Its derivative requires the chain rule, giving \(\partial w/\partial x = -e^{y-x}\).

Step 3: Apply the product rule and sum the results

The derivative of \( g(x,y) = u(x,y)*v(x,y)*w(x,y) \) with respect to 'x' can be found using the product rule, which states that the derivative of two multiplied functions is the first times the derivative of the second, plus the second times the derivative of the first. In this case, applying the rule repeatedly: \n\(\partial g/\partial x = u'vw + uv'w + uvw'\) \nSubstituting the differentiated values from step 2: \n\(\partial g/\partial x = 3*y^2*e^{y-x} + 0 + 3x*y^2*(-e^{y-x}) = 3y^2e^{y-x}(1 - x)\)

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus that helps us differentiate functions that are the product of two or more sub-functions. It gives us a way to find the derivative of a product without having to multiply the functions first. This is particularly useful in multivariable calculus when dealing with functions that depend on more than one variable.
This exercise demonstrates the application of the product rule to the function \(g(x, y) = 3xy^2e^{y-x}\). The product rule states that if you have a product of functions \(u(x, y), v(x, y),\) and \(w(x, y)\), then the derivative of the product \(uv\) with respect to \(x\) is:

• \((uv)' = u'v + uv'\)

In our case, the function \(g(x, y)\) is already expressed as three sub-functions \(u(x,y) = 3x\), \(v(x,y) = y^2\), and \(w(x,y) = e^{y-x}\). The product rule for three functions can be extended as:

• \(uvw' = u'vw + uv'w + uvw'\)

Here, we identify each sub-function and use their derivatives to find \(\partial g/\partial x\). This requires the individual derivatives of each sub-function, which are computed in the solution.
Breaking down the function with the product rule helps simplify the process and ensures that we can accurately determine the derivative of complex multivariable functions.

Chain Rule

The chain rule is another powerful tool in calculus that allows us to differentiate functions composed of nested sub-functions. It is particularly handy when dealing with exponential functions, as seen in the given problem. In this exercise, the third sub-function \(w(x, y) = e^{y-x}\) involves an exponential term with a composite argument \((y-x)\). The chain rule helps by allowing us to differentiate the outer function while also accounting for the derivative of the inner function.
For a function \(w(x, y) = e^{y-x}\), applying the chain rule involves taking the derivative of the outer function \(e^u\) where \(u = y-x\). The derivative of \(e^u\) with respect to \(u\) is \(e^u\). However, because \(u\) is itself a function of both \(x\) and \(y\), we calculate \(\partial u/\partial x = -1\). Therefore, through the chain rule, the derivative \(\partial w/\partial x\) becomes:

• \(-e^{y-x}\)

By applying the chain rule, we carefully account for changes in the composition of functions, allowing us to accurately compute complex derivatives that involve nested functions.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of two or more variables. It involves techniques that are crucial for analyzing how functions behave when more than one individual input affects the outcome. In essence, it allows for the computation of partial derivatives, as seen in this exercise.
When dealing with a function like \(g(x, y) = 3xy^2e^{y-x}\), understanding partial derivatives is key. A partial derivative measures how a function changes as one of the variables changes while keeping others constant, offering insights into the slope or rate of change along a specific direction.
In this exercise, finding \(g_x(x, y)\) means calculating the derivative of \(g(x, y)\) concerning \(x\) only and treating \(y\) as a constant. The need to identify sub-functions \(((u(x,y), v(x,y), w(x,y)))\) and use rules such as the product rule and the chain rule, highlights the complexity of working in multiple dimensions.
Multivariable calculus is foundational for many advanced topics and applications, such as in physics and engineering, providing the necessary tools for studying real-world systems where multiple factors are at play.