Unlocking Derivatives: Step-by-Step Guide & Examples

Hey everyone! Today, we're diving into the fascinating world of derivatives, a core concept in calculus. We'll specifically tackle finding the derivative of a function like y = (3x + 1)³(4x + 5)⁵. Don't worry if it sounds intimidating at first; we'll break it down step by step, making it super clear and easy to follow. Derivatives are all about understanding how a function changes, and they're super useful in tons of real-world applications, from physics to economics. So, let's get started and demystify this important topic together! We will explore the chain rule, product rule, and power rule to successfully solve this derivative problem.

Understanding the Basics: Derivatives Explained

Alright, before we jump into the nitty-gritty of the problem, let's quickly recap what a derivative actually is. In simple terms, the derivative of a function tells us the rate at which the function's output (y-value) changes with respect to its input (x-value). Think of it as the slope of the tangent line to the function at any given point. This slope essentially describes how the function is trending - is it going up, going down, or staying flat? The process of finding the derivative is called differentiation. There are several rules that we can apply to the differentiation process. These rules allow us to find the derivative of complex functions without doing all the hard work from first principles. It is important to know the derivatives of basic functions such as constant, power, exponential, logarithmic, and trigonometric functions to master the differentiation process. Now, let's talk about the specific rules we'll need for our example. We will use the product rule and chain rule to solve this problem. The product rule helps us differentiate the product of two functions, while the chain rule helps us differentiate a composite function.

So, why is this important? Well, derivatives are fundamental tools in calculus and are used in a variety of fields. In physics, derivatives help us calculate velocity and acceleration from position functions. In economics, they're used to find marginal cost and revenue. Basically, derivatives are everywhere. Understanding derivatives is crucial for anyone studying math, science, engineering, or even economics. They provide a powerful way to analyze change and make predictions. Learning how to find derivatives will unlock your understanding of calculus and make other concepts easier to grasp. So, let's get into the specifics of our problem now!

Applying the Product Rule: Breaking Down the Problem

Okay, let's get down to business and figure out how to find the derivative of y = (3x + 1)³(4x + 5)⁵. Because our function is a product of two other functions, we will start with the product rule. The product rule states that if we have a function y = u(x)v(x), then its derivative is given by:

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

Here, u(x) = (3x + 1)³ and v(x) = (4x + 5)⁵. So, we need to find the derivatives of both u(x) and v(x). To do this, we're going to need another important rule called the chain rule. The chain rule is essential when dealing with composite functions (functions within functions). It tells us how to differentiate a function like f(g(x)). In this case, our functions u(x) and v(x) are both composite functions.

First, let's find the derivative of u(x) = (3x + 1)³. We'll use the chain rule here. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In other words, differentiate the outer function first, keeping the inner function as it is, and then multiply by the derivative of the inner function. For u(x), the outer function is something cubed, and the inner function is (3x + 1). So, applying the chain rule:

• u'(x) = 3(3x + 1)² * (3) = 9(3x + 1)². We found the derivative of the outer function which is 3(3x+1)^2 and then multiplied by the derivative of the inner function (3x+1) which is 3.

Now, let's find the derivative of v(x) = (4x + 5)⁵, applying the same chain rule: v'(x) = 5(4x + 5)⁴ * (4) = 20(4x + 5)⁴. Similarly, we found the derivative of the outer function which is 5(4x+5)^4 and multiplied by the derivative of the inner function (4x+5) which is 4.

Now that we have u'(x) and v'(x), we're ready to plug everything back into the product rule formula and wrap this up!

The Chain Rule: Differentiating Composite Functions

As we saw earlier, the chain rule is a lifesaver when dealing with composite functions. Remember, the chain rule allows us to differentiate a function within another function. The chain rule is one of the most important rules in calculus and is used extensively.

Let's refresh our understanding of the chain rule with a couple of examples. Consider a function such as y = sin(2x). Here, the outer function is sin( ) and the inner function is 2x. To find the derivative, we first differentiate the outer function (sin), keeping the inner function as it is. The derivative of sin(x) is cos(x). So, we have cos(2x). Then we multiply this by the derivative of the inner function, which is 2. The final result is dy/dx = 2cos(2x).

Let's try another example. Consider y = (x² + 3)⁴. The outer function is something to the power of 4, and the inner function is x² + 3. Differentiating the outer function gives us 4(x² + 3)³. Multiplying this by the derivative of the inner function (2x), we get dy/dx = 8x(x² + 3)³.

Mastering the chain rule is all about practice. By identifying the outer and inner functions and applying the rule consistently, you will become comfortable with it. The chain rule can be applied multiple times in more complex scenarios where there are multiple nested functions. Therefore, practicing chain rule problems is a good way to understand the concepts and apply them to complex scenarios. This will make differentiating more complex functions easier.

Putting It All Together: Final Calculation

Okay, here we go, the grand finale! We have all the pieces we need to find the derivative of our original function y = (3x + 1)³(4x + 5)⁵. We will now use the product rule formula:

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

We know:

• u'(x) = 9(3x + 1)²*
• v(x) = (4x + 5)⁵*
• u(x) = (3x + 1)³*
• v'(x) = 20(4x + 5)⁴*

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

dy/dx = [9(3x + 1)² * (4x + 5)⁵] + [(3x + 1)³ * 20(4x + 5)⁴]. Now, we can try to simplify this expression a bit to make it look a little cleaner. We can factor out common terms, such as (3x+1)^2 and (4x+5)^4.

dy/dx = (3x + 1)²(4x + 5)⁴[9(4x + 5) + 20(3x + 1)]

Now, simplify the terms inside the brackets:

dy/dx = (3x + 1)²(4x + 5)⁴[36x + 45 + 60x + 20]. Collecting like terms we get:

dy/dx = (3x + 1)²(4x + 5)⁴(96x + 65)

And there you have it! The derivative of y = (3x + 1)³(4x + 5)⁵ is dy/dx = (3x + 1)²(4x + 5)⁴(96x + 65). Now, you’ve not only solved the problem, but you also have a solid understanding of how we got there. Congratulations!

Conclusion: Mastering Derivatives

Awesome work, guys! We've successfully found the derivative of a pretty complex function. We used a combination of the product rule and chain rule and some simple algebraic simplification, making this whole process a lot easier to digest. Remember, practice is key. The more you work through problems like these, the more comfortable and confident you'll become. Keep practicing and applying these rules, and you'll be well on your way to mastering calculus. Now you should be able to approach other similar problems with confidence. Remember to always look at the structure of the function and choose the appropriate differentiation rules. Good luck and keep learning!