Calculus: Finding The Derivative Of F(x) = X²e^(x-1)

Hey math enthusiasts! Let's dive headfirst into the world of calculus and tackle a fun problem. Today, we're going to use the rules of differentiation to find the derivative of the function f(x) = x²e^(x-1). Don't worry if things seem a bit tricky at first; we'll break it down step-by-step, making sure it's all crystal clear. This is a classic example that beautifully showcases the power of the product rule and chain rule in calculus. So, grab your pencils and get ready to learn! Understanding derivatives is super important in many areas, from physics to economics, so getting a solid grasp of these concepts is definitely worth the effort. Ready to roll?

Understanding the Basics: Derivatives and Rules

Before we jump into the problem, let's quickly recap some fundamental concepts. The derivative of a function, denoted as f'(x), essentially tells us the instantaneous rate of change of the function at any given point. Think of it as the slope of the tangent line to the function's graph at that specific point. It's like zooming in on a curve and finding how steep it is at a particular spot. In our case, we're trying to find how the function f(x) = x²e^(x-1) changes as x changes. The tools we'll be using are the product rule and the chain rule. These are like our secret weapons in this calculus adventure.

• The Product Rule: This rule comes into play when we have a function that's a product of two other functions. If f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Basically, it says that the derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. Simple, right?
• The Chain Rule: The chain rule is our go-to when we have a composite function, that is, a function within a function. If f(x) = u(v(x)), then f'(x) = u'(v(x)) * v'(x). We take the derivative of the outer function, evaluated at the inner function, and then multiply it by the derivative of the inner function. It's like peeling back the layers of an onion, one derivative at a time. The chain rule is the workhorse of differentiation, and we will definitely need it here.

Got it? Great! If not, don’t sweat it. We'll be using these rules directly, so you'll see how they work in action. The best way to understand these concepts is by doing. So, let’s get our hands dirty with the actual problem!

Applying the Product Rule to Our Function

Alright, let's apply these rules to our function, f(x) = x²e^(x-1). Notice that our function is the product of two functions: u(x) = x² and v(x) = e^(x-1). So, we'll need to use the product rule. First, we need to find the derivatives of u(x) and v(x).

• Finding u'(x): The derivative of u(x) = x² is pretty straightforward. Using the power rule (the derivative of x^n is n*x^(n-1)), we get u'(x) = 2x. Easy peasy!
• Finding v'(x): This is where things get slightly more interesting. We have v(x) = e^(x-1). Here, we'll need the chain rule because we have a function within a function: the exponential function with the exponent (x-1). The derivative of e^x is e^x, so the derivative of the outer function (the exponential) is still e to the power of whatever is inside. Then, we multiply by the derivative of the inner function, which is the derivative of (x-1). The derivative of (x-1) is simply 1, since the derivative of x is 1, and the derivative of a constant is 0. Therefore, v'(x) = e^(x-1) * 1 = e^(x-1).

Now that we have u'(x) and v'(x), we can plug them into the product rule formula: f'(x) = u'(x)v(x) + u(x)v'(x). Let's do it!

Putting It All Together: Calculating f'(x)

Okay, guys, we’re at the exciting part – the grand finale! We have all the pieces we need to calculate f'(x). Let's recap what we've found:

• u(x) = x²
• v(x) = e^(x-1)
• u'(x) = 2x
• v'(x) = e^(x-1)

Now, let's plug these into the product rule formula: f'(x) = u'(x)v(x) + u(x)v'(x).

• f'(x) = (2x)(e^(x-1)) + (x²)(e^(x-1))

And there you have it! That's the derivative of f(x) = x²e^(x-1). We've used the product rule and chain rule to find it. This form is perfectly valid, and we were instructed not to simplify, so we're done here. Pretty neat, huh?

Now, let's just make the presentation a little more tidy. You can also factor out the e^(x-1) if you want, but that wasn’t a requirement here. The following is also correct:

• f'(x) = e^(x-1)(2x + x²)

Both answers are correct. It just depends on what you want to do with the answer. We've managed to navigate through the product and chain rules to get to the answer. Awesome stuff!

Further Exploration and Practice

Congratulations on making it this far, everyone! You've successfully found the derivative of f(x) = x²e^(x-1). But the journey doesn't end here! Practice is key to mastering calculus, so I highly recommend you try similar problems to solidify your understanding. Here are a few suggestions to take your skills to the next level:

1. Try Different Functions: Experiment with other functions that involve products and composite functions. For instance, try finding the derivative of f(x) = x³sin(x), f(x) = (x²+1)e^(2x), or f(x) = ln(x² + 1). The more functions you work with, the more comfortable you'll become with applying the product rule and chain rule.
2. Simplify and Analyze: Once you've found the derivative, try simplifying the expression if possible. This can help you understand the behavior of the derivative and identify critical points where the function's rate of change might be interesting. For example, can you figure out where the slope of the tangent line is zero? Where is the function increasing and decreasing?
3. Use Online Tools: There are tons of online derivative calculators and resources. Use them to check your answers and get step-by-step solutions. This is an awesome way to learn from your mistakes and see how others approach the same problems. Websites like Wolfram Alpha are amazing for this.
4. Connect to Real-World Applications: Think about how derivatives are used in the real world. For instance, in physics, the derivative of displacement is velocity, and the derivative of velocity is acceleration. In economics, derivatives can be used to calculate marginal cost and marginal revenue. Understanding these applications can give you a better appreciation for the power and usefulness of calculus.

By practicing regularly and exploring different types of problems, you'll become more confident in your calculus abilities. Keep in mind that math is all about persistence and practice. Every time you solve a problem, you get a little bit better and a little bit closer to mastering the material. So, keep up the great work, and don't be afraid to ask for help if you get stuck. Your efforts will surely pay off!

Conclusion: Mastering the Derivative

Well, folks, we've successfully navigated the process of finding the derivative of f(x) = x²e^(x-1). We've seen how to apply the product rule and chain rule in a step-by-step manner, making sure that everything is easy to digest. Remember that practice is super important, so try out more problems to hone your skills. Keep exploring, keep learning, and don't be afraid to challenge yourself. Calculus is a beautiful and powerful tool, and with a little bit of effort, you can master it.

So, until next time, keep those math muscles flexing. I hope this was helpful. If you have any more questions or want to explore other calculus concepts, let me know. Happy deriving, everyone! Keep up the great work, and remember, you've got this!