Calculate The Rate Of Change: A Step-by-Step Guide

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Hey math enthusiasts! Today, we're diving into a fun little problem: finding the rate of change of a function at a specific point. Specifically, we're going to figure out the rate of change for the function f(x)=oot3ln(4−x2)f(x) = oot{3}\\{ln(4-x^2)} at x=1x = 1. Don't worry, it's not as scary as it sounds! We'll break it down into manageable steps, making sure everyone can follow along. Ready? Let's get started!

Understanding the Rate of Change

First things first, what exactly does "rate of change" mean, right? Essentially, it's a way to describe how much the output of a function (f(x)f(x) or the y-value) changes as the input (xx-value) changes. Think of it like this: If you're driving a car, the rate of change of your position is your speed. If your position changes a lot in a short amount of time, you're going fast (high rate of change); if your position doesn't change much, you're going slow (low rate of change). In the world of calculus, the rate of change is more precisely defined by the derivative of the function. The derivative gives us the instantaneous rate of change at any specific point.

So, when we're asked to find the rate of change at x=1x = 1, we're really being asked to find the value of the derivative of the function f(x)f(x) at x=1x = 1. This involves using the rules of differentiation, which help us find the derivative of a wide variety of functions. In other words, to find the rate of change of a function, you calculate its derivative. The derivative will give you another function, and when you plug in a specific x value (like x=1x = 1), you'll get the rate of change at that point. Let's get into the specifics of our function. The function in question, f(x)=oot3ln(4−x2)f(x) = oot{3}\\{ln(4-x^2)}, is a composition of several functions. This means we'll need to use the chain rule, one of the most important tools in calculus, to find its derivative. The chain rule helps us differentiate composite functions (functions within functions). It's a lifesaver when you're dealing with these kinds of expressions. Without the chain rule, calculating derivatives of complex functions would be a nightmare. We'll walk through this step by step, so even if you're new to this, you'll be able to follow along. Just remember that the chain rule says the derivative of a composite function is the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function. Sounds complicated? It isn't, really! Let's get to work!

Step-by-Step Calculation: Finding the Derivative

Alright, let's roll up our sleeves and actually do this. To find the rate of change, we need to find the derivative, f′(x)f'(x). Our function is f(x)=oot3ln(4−x2)f(x) = oot{3}\\{ln(4-x^2)}. First, let's rewrite this using exponents to make it easier to differentiate: f(x)=(ln(4−x2))13f(x) = (ln(4-x^2))^{\frac{1}{3}}. Now, we apply the chain rule. This rule states that if we have a function within a function (like we do here), we take the derivative of the outer function, leaving the inner function as it is, and then multiply by the derivative of the inner function.

  1. Outer Function: The outer function is the cube root, or the power of 13\frac{1}{3}. The derivative of u13u^{\frac{1}{3}} is 13u−23\frac{1}{3}u^{\frac{-2}{3}}. In our case, u is ln(4−x2)ln(4-x^2). So, we'll get 13(ln(4−x2))−23\frac{1}{3}(ln(4-x^2))^{\frac{-2}{3}}.
  2. Inner Function: The inner function is ln(4−x2)ln(4-x^2). We need to find its derivative. This also requires the chain rule! The derivative of ln(v)ln(v) is 1v∗v′\frac{1}{v} * v', where $v' $ is the derivative of v. Here, v=4−x2v = 4 - x^2, so v′=−2xv' = -2x. Therefore, the derivative of ln(4−x2)ln(4-x^2) is 14−x2∗−2x\frac{1}{4-x^2} * -2x, which simplifies to −2x4−x2\frac{-2x}{4-x^2}.
  3. Putting it Together: Now, we multiply the derivative of the outer function by the derivative of the inner function: f′(x)=13(ln(4−x2))−23∗−2x4−x2f'(x) = \frac{1}{3}(ln(4-x^2))^{\frac{-2}{3}} * \frac{-2x}{4-x^2}.

This is the derivative of our function f(x)f(x). Now, we're almost there! We've found the general formula for the rate of change at any x value. The next step is to plug in x=1x = 1 to find the rate of change specifically at that point. Remember, the derivative f′(x)f'(x) gives us the instantaneous rate of change for any x value. Let's do this to get a final solution. Notice that we applied the power rule and the chain rule to deal with the derivative of our function. These are two of the most fundamental rules of calculus, so understanding them is crucial.

Calculate the Rate of Change at x = 1

Now for the grand finale! We've got the derivative f′(x)=13(ln(4−x2))−23∗−2x4−x2f'(x) = \frac{1}{3}(ln(4-x^2))^{\frac{-2}{3}} * \frac{-2x}{4-x^2}, and we want to find the rate of change at x=1x = 1. This simply means we substitute 1 for every instance of x in the derivative equation. Let's do it.

  1. Substitute x=1x = 1: f′(1)=13(ln(4−(1)2))−23∗−2(1)4−(1)2f'(1) = \frac{1}{3}(ln(4-(1)^2))^{\frac{-2}{3}} * \frac{-2(1)}{4-(1)^2}.
  2. Simplify: f′(1)=13(ln(4−1))−23∗−24−1f'(1) = \frac{1}{3}(ln(4-1))^{\frac{-2}{3}} * \frac{-2}{4-1}.
  3. Further simplification: f′(1)=13(ln(3))−23∗−23f'(1) = \frac{1}{3}(ln(3))^{\frac{-2}{3}} * \frac{-2}{3}.
  4. Approximately evaluate: f′(1)≈13(1.099)−23∗−23≈13∗0.87∗−23≈−0.193f'(1) ≈ \frac{1}{3}(1.099)^{\frac{-2}{3}} * \frac{-2}{3} ≈ \frac{1}{3} * 0.87 * \frac{-2}{3} ≈ -0.193

So, the rate of change of the function f(x)=\root3ln(4−x2)f(x) = \root{3}\\{ln(4-x^2)} at x=1x = 1 is approximately −0.193-0.193. This means that at the point where x equals 1, the function's y-value is decreasing at a rate of approximately 0.193 units for every 1-unit increase in x. Neat, right?

The negative sign tells us that the function is decreasing at that point. If we had gotten a positive value, it would mean the function was increasing. It's really that simple! Always remember that the derivative gives us the slope of the tangent line to the function at a particular point, which represents the instantaneous rate of change. We used the chain rule twice here, once for the cube root and once inside the natural logarithm. The more practice you get with these rules, the easier this process will become. Also, using a calculator for the final numerical approximation is perfectly acceptable; the important thing is understanding the process of finding the derivative and substituting values.

Conclusion

Alright, guys, we did it! We successfully found the rate of change of the given function at x=1x = 1. We broke down a seemingly complex problem into a series of understandable steps. We learned about the rate of change, the derivative, and the chain rule, and we applied these concepts to solve the problem. Remember, the key to mastering calculus is practice. Work through different examples, get comfortable with the rules, and don't be afraid to ask for help when you need it. Math can be challenging, but it's also incredibly rewarding when you finally "get" it. Keep practicing, and you'll find it gets easier and more intuitive! Well done, and keep exploring the amazing world of mathematics! Calculus is a powerful tool with applications across many fields, from physics and engineering to economics and computer science. Keep up the excellent work, and enjoy the journey!