Unlocking Derivatives: Solving K'(2) Step-by-Step

by Andrew McMorgan 50 views

Hey Plastik Magazine readers! Ever stumbled upon a math problem and thought, "Whoa, where do I even begin?" Well, today, we're diving into a calculus problem that might seem a bit intimidating at first glance: finding the derivative of a function and evaluating it at a specific point. Don't worry, though; we'll break it down step by step, making it super easy to understand. So, grab your coffee, and let's get started. We are going to find K'(2), where K(x) = 3 + 3x^2 + e^(2x-4). This is a common problem in calculus, and understanding the process will help you tackle many other derivative-related challenges, so pay attention, guys!

Decoding the Problem: What Does K'(2) Mean?

Before we jump into the solution, let's make sure we're all on the same page. The notation K'(2) represents the derivative of the function K(x) evaluated at x = 2. In simpler terms, we need to find the instantaneous rate of change of the function K(x) at the point where x equals 2. The derivative tells us how the function's output (y-value) changes concerning a tiny change in the input (x-value). Think of it like this: If K(x) represents the position of a car at time x, then K'(x) represents the car's velocity at time x, and K'(2) represents the car's velocity at time equals 2. Pretty cool, huh? This concept is fundamental to calculus and has applications in various fields, from physics and engineering to economics and computer science. Remember that the derivative is another function derived from the original function. We first need to compute K'(x), then we evaluate this at x=2. So, how do we find K'(x)? Let's go through the necessary steps.

Now, let's explore how to solve this step-by-step. The process is broken down into manageable parts. We'll start by taking the derivative of each term in the function K(x) and then evaluate the result at the specified point. This method allows us to dissect the problem and grasp each component individually, making the entire process easier to comprehend and more enjoyable. Let's get to work!

Step 1: Find the Derivative of K(x), K'(x)

Alright, guys, let's get down to business. To find K'(2), the first thing we need to do is calculate the derivative of the function K(x). Recall that K(x) = 3 + 3x^2 + e^(2x-4). We'll use the power rule and the chain rule, which are two essential tools in your calculus toolkit. Let's break down each term of the function and find its derivative. First, we have the constant term, 3. The derivative of a constant is always zero. This is because a constant doesn't change; its rate of change is zero. Next, we have the term 3x^2. To find its derivative, we'll use the power rule, which states that if you have a term like ax^n, its derivative is n*ax^(n-1). Applying the power rule to 3x^2, we get 2 * 3x^(2-1), which simplifies to 6x. Lastly, we have the term e^(2x-4). This requires the chain rule because we have a function inside another function (e to the power of something). The chain rule states that the derivative of f(g(x)) is f'(g(x)) * g'(x). In our case, f(u) = e^u and g(x) = 2x-4. The derivative of e^u is e^u, and the derivative of 2x-4 is 2. Therefore, the derivative of e^(2x-4) is e^(2x-4) * 2 or 2e^(2x-4). Now, let's put it all together. So, after having computed all the derivatives of the individual terms, the derivative of K(x), which we denote as K'(x), is equal to 0 + 6x + 2e^(2x-4), which is equivalent to 6x + 2e^(2x-4). Congratulations, guys, we have already done the first and most challenging part!

To make it clearer, here is the breakdown:

  • Derivative of 3: 0
  • Derivative of 3x^2: 6x
  • Derivative of e^(2x-4): 2e^(2x-4)
  • Therefore, K'(x) = 6x + 2e^(2x-4)

Step 2: Evaluate K'(x) at x = 2

We're in the home stretch, guys! Now that we've found the derivative K'(x) = 6x + 2e^(2x-4), all that's left is to evaluate it at x = 2. This means we substitute every instance of 'x' in our derivative equation with the value '2'. Let's do it! So, K'(2) = 6 * (2) + 2e^(2*(2) - 4). First, simplify the terms inside the exponential function: 2 * (2) - 4 = 4 - 4 = 0. Therefore, our expression now looks like this: K'(2) = 6 * (2) + 2e^(0). Now, evaluate each term. We know that 6 * 2 = 12. Also, recall that any number raised to the power of 0 equals 1. Therefore, e^(0) = 1. So now, our equation is: K'(2) = 12 + 2 * 1. And finally, 12 + 2 * 1 = 12 + 2 = 14. So, K'(2) = 14. The final result represents the instantaneous rate of change of the function K(x) at the point where x = 2. This tells us how fast the function is changing at that specific point. It’s the slope of the tangent line to the function's curve at x = 2. Understanding this concept is important in many areas of mathematics and science.

Now, let's summarize the calculation steps:

  • K'(x) = 6x + 2e^(2x-4)
  • Substitute x = 2: K'(2) = 6 * (2) + 2e^(2*(2) - 4)
  • Simplify: K'(2) = 12 + 2e^(0)
  • Calculate: K'(2) = 12 + 2 = 14

Step 3: Conclusion

And there you have it, guys! We've successfully found K'(2), and the answer is 14. You've now seen how to calculate the derivative of a function and evaluate it at a specific point. Remember that practice is key, so don't hesitate to work through more examples. As you become more comfortable with these steps, you'll find that calculus becomes less intimidating and more of an exciting puzzle to solve. Calculus is fundamental to so many areas of science, technology, engineering, and mathematics. Knowing how to find derivatives is critical in these areas. Keep up the great work, and happy calculating!

This problem showed us the basics of finding a derivative and evaluating it at a point. Remember the most important rules, like the power rule and the chain rule. You should also remember how to simplify exponents. Practicing these problems will help you understand calculus better. Now you are one step closer to mastering calculus!