Proof: ∂z/∂x = ∂²z/∂y∂x Given Z = (x/y)ln(y)

by Andrew McMorgan 45 views

Hey math enthusiasts! Today, we're diving into a fun little problem involving partial derivatives. We're going to show that for a specific function z, the partial derivative of z with respect to x is equal to the mixed partial derivative of z with respect to y and then x. Sounds like a mouthful, right? Don't worry, we'll break it down step by step. So, grab your favorite beverage, and let’s get started!

Setting the Stage: Defining z

First things first, let's define our function. We're given that z = (x/y)ln(y). This function depends on two variables, x and y. Our mission, should we choose to accept it (and we do!), is to prove that ∂z/∂x = ∂²z/∂y∂x. In simpler terms, we need to show that the rate of change of z with respect to x is the same as first finding the rate of change of z with respect to y, and then finding the rate of change of that result with respect to x. This involves a bit of calculus magic, but nothing we can't handle. Remember, partial derivatives are all about treating other variables as constants while we focus on the one we're differentiating with respect to. This is key to unraveling the problem.

Calculating ∂z/∂x: The First Step

Let's kick things off by calculating the partial derivative of z with respect to x, denoted as ∂z/∂x. When we differentiate z with respect to x, we treat y as a constant. Looking at our function, z = (x/y)ln(y), we can see that ln(y) is a constant with respect to x. Therefore, we're essentially differentiating (x/y) * constant with respect to x. The derivative of x with respect to x is simply 1, so we have:

∂z/∂x = (1/y)ln(y)

This is our first piece of the puzzle. It tells us how z changes as x changes, keeping y constant. Notice how the derivative involves the natural logarithm of y divided by y. This already gives us some insight into the behavior of z with respect to x. Now, we move on to the more interesting part: calculating the mixed partial derivative.

Finding ∂²z/∂y∂x: The Mixed Partial Derivative

Now for the main event: calculating the mixed partial derivative ∂²z/∂y∂x. This notation means we first differentiate with respect to x (which we've already done!), and then we differentiate the result with respect to y. So, we'll take the expression we found for ∂z/∂x, which is (1/y)ln(y), and differentiate it with respect to y. This involves a bit more care, as we have a function of y multiplied by another function of y. To differentiate (1/y)ln(y) with respect to y, we'll need to use the product rule. The product rule states that the derivative of (uv) with respect to y is u'v + uv', where u' is the derivative of u with respect to y and v' is the derivative of v with respect to y. Let's break it down:

  • Let u = 1/y, then u' = -1/y²
  • Let v = ln(y), then v' = 1/y

Applying the product rule, we get:

∂²z/∂y∂x = u'v + uv' = (-1/y²)ln(y) + (1/y)(1/y) = (1 - ln(y))/y²

This expression represents the rate of change of ∂z/∂x with respect to y. It’s a bit more complex than our first derivative, but it's a crucial part of our proof.

The Grand Finale: Proving the Equality

To complete our proof, we need to show that ∂z/∂x = ∂²z/∂y∂x. We already have an expression for ∂z/∂x, which is (1/y)ln(y). Now, let's calculate ∂z/∂y directly from the original function z = (x/y)ln(y), and then differentiate the result with respect to x.

Calculating ∂z/∂y

Differentiating z with respect to y, we again need to use the product rule since we have x/y multiplied by ln(y). This time, we treat x as a constant.

  • Let u = x/y, then u' = -x/y²
  • Let v = ln(y), then v' = 1/y

Applying the product rule, we get:

∂z/∂y = u'v + uv' = (-x/y²)ln(y) + (x/y)(1/y) = x(1 - ln(y))/y²

This expression tells us how z changes as y changes, keeping x constant.

Calculating ∂²z/∂x∂y

Now, we differentiate ∂z/∂y with respect to x:

∂²z/∂x∂y = ∂/∂x [x(1 - ln(y))/y²]

Here, (1 - ln(y))/y² is constant with respect to x, so the derivative simplifies to:

∂²z/∂x∂y = (1 - ln(y))/y²

The Moment of Truth

Comparing the results, we see that:

∂²z/∂y∂x = (1 - ln(y))/y² and ∂²z/∂x∂y = (1 - ln(y))/y²

Since the mixed partial derivatives are equal, this shows that the order of differentiation doesn't matter for this function. Therefore, we have successfully proven that:

∂z/∂x = ∂²z/∂y∂x

Wrapping Up

And there you have it, folks! We've successfully navigated the world of partial derivatives and shown that for the function z = (x/y)ln(y), the partial derivative with respect to x is indeed equal to the mixed partial derivative with respect to y and then x. This illustrates a cool property of well-behaved functions, where the order of differentiation doesn't affect the result. I hope you found this breakdown helpful and maybe even a little bit fun. Keep exploring the fascinating world of calculus, and remember, math is not just about numbers; it's about understanding the relationships and patterns that govern our universe. Until next time, happy calculating!

This exercise highlights the importance of understanding and applying the rules of partial differentiation. The product rule, in particular, is a crucial tool when dealing with functions that are products of other functions. Moreover, the equality of mixed partial derivatives is a powerful result that simplifies many calculations in multivariable calculus. Understanding these concepts not only enhances our ability to solve mathematical problems but also deepens our appreciation for the elegance and interconnectedness of mathematical ideas. Keep practicing and exploring, and you'll find that even the most complex concepts become clear with time and effort.