Differentiating Complex Functions: A Calculus Guide

Unlock the Secrets of Calculus: Differentiating Complex Functions

Hey guys, are you ready to dive into the fascinating world of calculus? Today, we're going to tackle a pretty cool problem: differentiating the function $y=e^{\ln (3 x+2)} \cos \left[\frac{3 x+2}{4 x}\right]$. This might look a bit intimidating at first glance, but trust me, we'll break it down step by step to make it super easy to understand. Our main goal here is to find the derivative of this function, which tells us how the function changes as the input variable x changes. Understanding derivatives is fundamental in calculus, opening doors to solving problems in physics, engineering, economics, and so much more. We'll use a combination of the chain rule, product rule, and some clever simplifications to find our answer. So, grab your coffee, get comfortable, and let's start our calculus adventure! This problem beautifully combines several key concepts, including exponential functions, logarithmic functions, and trigonometric functions, providing an excellent exercise in applying the rules of differentiation. As we work through this, we'll reinforce our understanding of these fundamental principles. The derivative of a function gives us the slope of the tangent line at any point on the function's graph. This is incredibly useful for finding things like the maximum or minimum values of a function, or determining the rate of change of a quantity. Remember, practice is key when it comes to calculus, so by the end of this, you'll be well on your way to mastering the art of differentiation. We’ll also emphasize the importance of simplifying the expression before diving into the differentiation process. Simplifying can significantly reduce the complexity of the problem and make the differentiation process much smoother. This approach not only makes the problem easier to solve but also helps in understanding the underlying mathematical principles more deeply. Let's get started! We'll begin by simplifying our function. Using the property that $e^{\ln(u)} = u$, we can simplify the first part of the function. This simplification is a great example of how understanding the properties of functions can make complex problems much more manageable. This step alone streamlines our function and makes it much easier to handle. Always look for these kinds of opportunities – they save time and reduce the chances of making mistakes. This simplification will also help us to keep things organized and prevent us from getting lost in the more complex calculations. We aim to ensure that the derivative is calculated correctly, and we understand each step in this process. Remember, every step in calculus builds on previous steps, so it's important to be methodical and precise. This approach is vital for both understanding and applying calculus effectively.

Breaking Down the Problem: Step-by-Step Differentiation

Okay, let's break down this differentiation problem step by step, shall we? First, remember our function is $y=e^{\ln (3 x+2)} \cos \left[\frac{3 x+2}{4 x}\right]$. As we mentioned before, the first thing we should do is simplify the expression using the property $e^{\ln(u)} = u$. This gives us $y = (3x + 2) \cos \left[\frac{3 x+2}{4 x}\right]$. See, much cleaner already! Now, to find the derivative, we'll need to use the product rule, which states that the derivative of a product of two functions, say $u(x)$ and $v(x)$, is $(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)$. In our case, $u(x) = 3x + 2$ and $v(x) = \cos \left[\frac{3 x+2}{4 x}\right]$. So, we have to find the derivatives of both $u(x)$ and $v(x)$. Differentiating $u(x) = 3x + 2$ is straightforward; $u'(x) = 3$. Now, let's find the derivative of $v(x) = \cos \left[\frac{3 x+2}{4 x}\right]$. This requires using the chain rule, which says that if we have a function $f(g(x))$, its derivative is $f'(g(x)) \cdot g'(x)$. Here, the outer function is $\cos(z)$ and the inner function is $z = \frac{3x+2}{4x}$. The derivative of $\cos(z)$ is $-\sin(z)$, so we have $-\sin \left[\frac{3 x+2}{4 x}\right]$. Now we need to find the derivative of the inner function $z = \frac{3x+2}{4x}$. To differentiate this, we can use the quotient rule, which states that the derivative of $\frac{f(x)}{g(x)}$ is $\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$. Let $f(x) = 3x + 2$ and $g(x) = 4x$. Then $f'(x) = 3$ and $g'(x) = 4$. Applying the quotient rule, we get $\frac{3 \cdot 4x - (3x + 2) \cdot 4}{(4x)^2} = \frac{12x - 12x - 8}{16x^2} = \frac{-8}{16x^2} = \frac{-1}{2x^2}$. Putting everything together, the derivative of $v(x)$ is $-\sin \left[\frac{3 x+2}{4 x}\right] \cdot \frac{-1}{2x^2} = \frac{1}{2x^2} \sin \left[\frac{3 x+2}{4 x}\right]$. Now that we have all the pieces, we can apply the product rule: $y' = u'(x)v(x) + u(x)v'(x) = 3 \cdot \cos \left[\frac{3 x+2}{4 x}\right] + (3x + 2) \cdot \frac{1}{2x^2} \sin \left[\frac{3 x+2}{4 x}\right]$. Simplifying gives us the final derivative. Phew! We've successfully differentiated the function. That wasn't so bad, was it?

Mastering the Techniques: Chain Rule, Product Rule, and Quotient Rule in Action

Let's dig a little deeper into the techniques we just used: the chain rule, product rule, and quotient rule. These are the bread and butter of differentiation, and understanding them is crucial for any calculus enthusiast. The chain rule is your go-to tool when you have a function inside another function. In our example, we had $\cos \left[\frac{3 x+2}{4 x}\right]$. Here, the outer function is the cosine function, and the inner function is the fraction. The chain rule allows us to differentiate these nested functions step by step. You differentiate the outer function first, leaving the inner function untouched, and then you multiply by the derivative of the inner function. This allows us to unravel complex functions into manageable parts. The product rule, which we used to differentiate our simplified function $y = (3x + 2) \cos \left[\frac{3 x+2}{4 x}\right]$, is essential when you're dealing with the product of two functions. The rule tells you how to find the derivative of the product. You take the derivative of the first function, multiply it by the second function, then add the first function multiplied by the derivative of the second function. It’s a systematic way to handle these combinations and is used frequently in calculus applications. Finally, the quotient rule is your friend when you have one function divided by another. In our example, we used it to differentiate the inner function $\frac{3x+2}{4x}$. The quotient rule allows you to find the derivative of a function that is the result of one function divided by another. It helps us handle fractions within functions. All these rules are fundamental tools in your calculus toolkit. By mastering them, you will become much more confident in tackling a wide range of differentiation problems. These rules are not just abstract concepts; they are practical tools that you will use repeatedly throughout your calculus journey. Make sure to practice, practice, practice – the more you use them, the more comfortable and proficient you'll become. Remember, the key to success in calculus is not just memorizing the rules, but understanding when and how to apply them.

Simplifying and Finalizing the Derivative: A Clear Path to the Solution

Alright, let's go over the final steps to make sure we have a clear and concise solution. Remember, after applying all our rules, we had $y' = 3 \cdot \cos \left[\frac{3 x+2}{4 x}\right] + (3x + 2) \cdot \frac{1}{2x^2} \sin \left[\frac{3 x+2}{4 x}\right]$. While this is technically correct, it can be written more neatly. Let's rewrite it as $y' = 3 \cos \left[\frac{3 x+2}{4 x}\right] + \frac{3x + 2}{2x^2} \sin \left[\frac{3 x+2}{4 x}\right]$. Now, you might ask, can we simplify this further? Well, not really. There are no common factors we can easily extract to simplify it more. So, this is your final derivative! Remember, the process is more important than the final answer. We've gone through each step methodically, explaining the rules and the logic behind them. The key here is the step-by-step approach, where we simplify the function, use the appropriate differentiation rules (chain, product, and quotient), and arrive at the final answer. We started with a complex function and, through simplification and strategic application of differentiation rules, we arrived at a manageable and accurate derivative. This is a testament to the power of calculus and the importance of understanding the fundamentals. It's about applying the right techniques in the right order. This problem showcased how the product rule and chain rule must be applied together, along with algebraic simplification. This showcases the versatility of calculus in handling complex mathematical expressions. Keep in mind that the key is not just to get the answer, but to understand the process. Go through your work again to make sure all calculations are accurate, and you haven't missed anything. Check the derivative using online tools or other methods if necessary to confirm the result. Great job, everyone! You've successfully differentiated a complex function using the chain rule, the product rule, and the quotient rule.