Finding The Third Derivative Of U = E^(xyz): A Step-by-Step Guide

Hey there, math enthusiasts! Ever stumbled upon a problem that looks like a jumble of letters and symbols? Today, we're diving deep into one such problem: finding the third-order mixed partial derivative of the function u = e^(xyz). Sounds intimidating? Don't worry, we'll break it down step by step, making it super easy to understand. So, grab your calculators and let's get started!

Understanding the Problem: Breaking Down d³u/dxdydz

Before we jump into the calculations, let's make sure we understand the notation. When you see d³u/dxdydz, it's math shorthand for a third-order mixed partial derivative. What does that mean? Essentially, we're taking the derivative of the function u three times, each time with respect to a different variable. The order matters here! d³u/dxdydz tells us we first differentiate with respect to z, then y, and finally x. Think of it as a carefully choreographed dance of differentiation. We're dealing with a function, u = e^(xyz), where e is Euler's number (approximately 2.71828), and x, y, and z are our variables. Our mission? To find out what happens when we differentiate this function three times in a specific order. This kind of problem pops up a lot in advanced calculus and is super useful in fields like physics and engineering for analyzing complex systems. Understanding these derivatives helps us grasp how the function u changes when we tweak its variables – a crucial skill for any budding scientist or engineer!

Step 1: Differentiate with Respect to z (∂u/∂z)

Okay, let's get our hands dirty with some actual differentiation! Our first task is to find the partial derivative of u with respect to z, denoted as ∂u/∂z. Remember, u = e^(xyz). When we differentiate with respect to z, we treat x and y as constants. Think of them as numbers hanging out for the ride. To differentiate e^(xyz) with respect to z, we'll use the chain rule. The chain rule basically says that if you have a function inside another function, you differentiate the outer function, keeping the inner function the same, and then multiply by the derivative of the inner function. In our case, the outer function is e^u (where u is a placeholder) and the inner function is xyz. The derivative of e^u is simply e^u, so the derivative of e^(xyz) is e^(xyz). Now, we need to multiply by the derivative of the inner function, xyz, with respect to z. The derivative of xyz with respect to z is just xy (since x and y are constants). Putting it all together, we get: ∂u/∂z = xy * e^(xyz). Ta-da! We've taken our first step. We've found the first partial derivative, and it's looking pretty neat. This result, xy * e^(xyz), is crucial because it sets the stage for the next round of differentiation. We're one step closer to finding that elusive third-order derivative!

Step 2: Differentiate with Respect to y (∂/∂y (∂u/∂z))

Alright, we've conquered the first derivative; now it's time to tackle the second! We're taking the result from the previous step, ∂u/∂z = xy * e^(xyz), and we're going to differentiate it with respect to y. This is written as ∂/∂y (∂u/∂z). Just like before, when we differentiate with respect to y, we treat x and z as constants. This means they're just along for the ride and won't change during this differentiation step. Now, looking at xy * e^(xyz), we can see that we have a product of two functions involving y: xy and e^(xyz). This means we'll need to use the product rule. The product rule states that the derivative of (uv) with respect to a variable is u'v + uv', where u' is the derivative of u and v' is the derivative of v. Let's identify our u and v: Let u = xy, then u' (the derivative of u with respect to y) is just x. Let v = e^(xyz). To find v', we again use the chain rule. The derivative of e^(xyz) with respect to y is xz * e^(xyz) (we multiply by xz because that's the derivative of the exponent xyz with respect to y). Now we plug everything into the product rule formula: u'v + uv' = (x) * (e^(xyz)) + (xy) * (xz * e^(xyz)) Simplifying this, we get: x * e^(xyz) + x²yz * e^(xyz). We can factor out an x * e^(xyz)* to make it look even cleaner: x * e^(xyz) * (1 + xyz). Boom! We've found our second partial derivative. This expression might look a bit complex, but we're making great progress. We're now ready for the final step, where we differentiate one more time!

Step 3: Differentiate with Respect to x (∂/∂x (∂²u/∂y∂z))

We've made it to the final leg of our journey! We're taking the result from the previous step, x * e^(xyz) * (1 + xyz), and differentiating it with respect to x. This gives us ∂/∂x (∂²u/∂y∂z), which is the third-order mixed partial derivative we've been aiming for. This step is a bit more involved, but don't worry, we'll break it down. Remember, when differentiating with respect to x, we treat y and z as constants. Looking at our expression, x * e^(xyz) * (1 + xyz), we can see it as a product of two functions: x and e^(xyz) * (1 + xyz). So, we'll need to use the product rule again. Let's identify our u and v: Let u = x, then u' (the derivative of u with respect to x) is simply 1. Let v = e^(xyz) * (1 + xyz). Finding v' is a little trickier. We'll need to use the product rule again within this derivative! Think of v as a product of two functions: e^(xyz) and (1 + xyz). Let's call these a and b for clarity. So, a = e^(xyz), and the derivative of a with respect to x (a') is yz * e^(xyz) (using the chain rule). And, b = (1 + xyz), so the derivative of b with respect to x (b') is yz. Now, using the product rule for v, we have: v' = a'b + ab' = (yz * e^(xyz)) * (1 + xyz) + (e^(xyz)) * (yz) We can factor out a yz * e^(xyz)*: v' = yz * e^(xyz) * (1 + xyz + 1) = yz * e^(xyz) * (2 + xyz) Phew! That was a derivative within a derivative. Now we have everything we need to apply the product rule to the original problem: ∂/∂x [x * e^(xyz) * (1 + xyz)] = u'v + uv' = (1) * [e^(xyz) * (1 + xyz)] + (x) * [yz * e^(xyz) * (2 + xyz)] Let's simplify this beast: = e^(xyz) * (1 + xyz) + xyz * e^(xyz) * (2 + xyz) We can factor out an e^(xyz): = e^(xyz) * [(1 + xyz) + xyz * (2 + xyz)] = e^(xyz) * [1 + xyz + 2xyz + x²y²z²] Finally, we combine like terms: = e^(xyz) * (1 + 3xyz + x²y²z²) And there you have it! We've found the third-order mixed partial derivative: d³u/dxdydz = e^(xyz) * (1 + 3xyz + x²y²z²).

Final Answer: d³u/dxdydz = e^(xyz) * (1 + 3xyz + x²y²z²)

Woohoo! We did it, guys! We successfully navigated the twists and turns of partial differentiation and arrived at our final answer: d³u/dxdydz = e^(xyz) * (1 + 3xyz + x²y²z²). That might look like a complex expression, but we built it step by step, using the chain rule and the product rule like pros. Remember, the key to mastering these kinds of problems is to break them down into smaller, manageable steps. Don't be intimidated by the notation; just take it one derivative at a time. This problem is a fantastic example of how calculus can be used to understand the behavior of functions in multiple dimensions. It's a skill that will serve you well in many areas of science and engineering. So, pat yourselves on the back, you've tackled a challenging problem and come out on top! Keep practicing, keep exploring, and most importantly, keep enjoying the beauty of mathematics! If you ever get stuck, just remember this step-by-step guide, and you'll be differentiating like a champ in no time. And remember, math isn't just about finding the right answer; it's about the journey of discovery and the thrill of solving a puzzle. So, keep challenging yourselves, and who knows what amazing mathematical feats you'll achieve next!