Derivative Of Y=x²e⁻⁸ˣ: A Step-by-Step Guide

by ADMIN 45 views

Hey guys! Today, we're diving into the world of calculus to tackle a pretty common type of problem: finding the derivative of a function. Specifically, we're going to figure out how to differentiate the function y = x²e⁻⁸ˣ. This might look intimidating at first glance, but don't worry! We'll break it down step by step, making sure you understand each part of the process. So, grab your pencils, and let's get started!

Understanding the Problem: Product Rule

Before we jump into the calculations, it's crucial to recognize what we're dealing with. Our function, y = x²e⁻⁸ˣ, is a product of two smaller functions: x² and e⁻⁸ˣ. This immediately tells us that we'll need to employ the product rule of differentiation. Remember, the product rule is your best friend when you're differentiating a function that's made up of two functions multiplied together.

So, what is the product rule? In simple terms, it states that the derivative of the product of two functions is equal to the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. Mathematically, if we have two functions, u(x) and v(x), then the derivative of their product, y = u(x)v(x), is given by:

dy/dx = u'(x)v(x) + u(x)v'(x)

Where u'(x) represents the derivative of u(x), and v'(x) represents the derivative of v(x). Got it? Great! Now, let's apply this to our specific problem.

Step 1: Identifying u(x) and v(x)

First things first, we need to clearly identify our u(x) and v(x). Looking at our function, y = x²e⁻⁸ˣ, it's pretty straightforward:

  • u(x) = x²
  • v(x) = e⁻⁸ˣ

Easy peasy, right? Now that we've identified our u and v, the next step is to find their derivatives. This is where the power rule and chain rule come into play.

Step 2: Finding the Derivatives u'(x) and v'(x)

Let's start with u(x) = x². This is a classic power rule scenario. The power rule states that if u(x) = xⁿ, then u'(x) = nxⁿ⁻¹. Applying this to our u(x), we get:

u'(x) = 2x²⁻¹ = 2x

So, the derivative of x² is simply 2x. That was a breeze! Now, let's move on to v(x) = e⁻⁸ˣ. This one is a bit more interesting because it involves the chain rule. The chain rule is used when we're differentiating a composite function (a function within a function). In this case, we have the exponential function eˣ with an inner function of -8x.

The chain rule states that if v(x) = f(g(x)), then v'(x) = f'(g(x)) * g'(x). In simpler terms, we take the derivative of the outer function (leaving the inner function as is) and then multiply it by the derivative of the inner function.

Applying this to v(x) = e⁻⁸ˣ, we get:

  • The derivative of the outer function, eˣ, is just eˣ. So, the derivative of e⁻⁸ˣ (leaving the inner function as is) is e⁻⁸ˣ.
  • The derivative of the inner function, -8x, is simply -8.

Therefore, using the chain rule, the derivative of v(x) = e⁻⁸ˣ is:

v'(x) = e⁻⁸ˣ * (-8) = -8e⁻⁸ˣ

Alright! We've successfully found both u'(x) and v'(x). We're halfway there! Now comes the fun part: plugging everything into the product rule formula.

Step 3: Applying the Product Rule

Remember the product rule formula? It's: dy/dx = u'(x)v(x) + u(x)v'(x). We've got all the pieces, so let's put them together. We know:

  • u(x) = x²
  • v(x) = e⁻⁸ˣ
  • u'(x) = 2x
  • v'(x) = -8e⁻⁸ˣ

Substituting these values into the product rule formula, we get:

dy/dx = (2x)(e⁻⁸ˣ) + (x²)(-8e⁻⁸ˣ)

Okay, we've got our derivative! But it looks a bit messy, doesn't it? The final step is to simplify it.

Step 4: Simplifying the Derivative

To simplify our derivative, dy/dx = (2x)(e⁻⁸ˣ) + (x²)(-8e⁻⁸ˣ), we can look for common factors. Notice that both terms have e⁻⁸ˣ and x in them. Let's factor those out:

dy/dx = e⁻⁸ˣ(2x - 8x²)

We can simplify it a little further by factoring out a 2x from the parentheses:

dy/dx = 2xe⁻⁸ˣ(1 - 4x)

And there you have it! We've successfully found and simplified the derivative of y = x²e⁻⁸ˣ. Our final answer is:

dy/dx = 2xe⁻⁸ˣ(1 - 4x)

Conclusion: Mastering Derivatives

Awesome job, guys! You've made it through the process of finding the derivative of a function using the product rule and chain rule. Remember, the key to mastering calculus is practice. The more problems you solve, the more comfortable you'll become with the different rules and techniques. Don't be afraid to tackle challenging problems, and always break them down into smaller, manageable steps.

In this case, we saw how identifying the product rule, applying the chain rule, and simplifying the result led us to the correct derivative. Keep these steps in mind as you encounter similar problems in the future.

So, keep practicing, keep exploring, and most importantly, keep learning! You've got this! And if you ever get stuck, remember there are plenty of resources available online and in textbooks to help you out. Happy differentiating!