Mastering The Product Rule: $f(x)=(9x^3+2)(5x^2-4)$ Derivative

Dec 3, 2025 by Admin 63 views

Hey there, calculus explorers! Ever stared at a complex function and thought, "Ugh, how do I even begin to find its derivative?" You're definitely not alone, guys. Derivatives can sometimes feel like solving a mystery novel, but once you've got the right tools, it becomes a thrilling adventure. Today, we're diving deep into one of the most fundamental and powerful tools in your calculus arsenal: the Product Rule. Specifically, we're going to break down how to find the derivative of a function like $f(x)=(9x^3+2)(5x^2-4)$ . This isn't just about getting the right answer; it's about understanding the process, building your confidence, and making sure you're ready for any calculus challenge that comes your way. This particular function looks like two separate polynomials multiplied together, which is a big hint that the Product Rule is going to be our best friend here. We'll go through it step by step, using a friendly, conversational tone so you can really grasp the concepts. So, grab your virtual pencils, and let's unlock the secrets of this derivative problem together. We’ll make sure you not only solve it but truly master the underlying principles of differentiation when dealing with products of functions. Ready? Let's roll!

This journey will illuminate not just the mechanical steps of applying the Product Rule but also the reasoning behind each move. Understanding the 'why' behind the 'how' is what truly elevates your mathematical prowess. Many students find themselves memorizing formulas without really grasping their application, and that's where things can get tricky. Our goal here is to make sure that by the end of this article, you'll be able to confidently identify when and how to apply the Product Rule to similar functions. We'll also address the specific format of the problem, where you need to fill in those mysterious blanks, which is a common way these questions are posed in exams and problem sets. So, whether you're a seasoned calculus veteran looking for a refresher or a newbie just getting started, this guide on finding the derivative of $f(x)=(9x^3+2)(5x^2-4)$ using the product rule is designed to provide immense value and clarity. Let's make complex calculus feel simple and intuitive!

What's the Product Rule Anyway, Guys? A Quick Refresher

Alright, before we tackle our specific function, let's make sure we're all on the same page about the Product Rule. Think of it this way: when you have two functions, say $u(x)$ and $v(x)$ , that are multiplied together to form a new function, $f(x) = u(x)v(x)$ , you can't just find the derivative of each one separately and multiply those derivatives. That's a common mistake, and it won't get you the right answer! Instead, we use the Product Rule, which states: The derivative of a product of two functions is the first function times the derivative of the second, plus the second function times the derivative of the first.

In mathematical notation, it looks like this:

$d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

Pretty neat, right? Let's break down those symbols:

$u(x)$ is your first function.
$v(x)$ is your second function.
$u'(x)$ is the derivative of your first function.
$v'(x)$ is the derivative of your second function.

It's like a dance! One function takes the lead while the other stays put, then they switch roles. You take the derivative of one, leave the other alone, and then add that to the first one staying put while you take the derivative of the second. This elegant formula ensures that we correctly account for how both functions contribute to the overall rate of change of their product. This rule is absolutely essential when you can't easily multiply out the functions before differentiating, such as with trigonometric functions or exponentials. It's a cornerstone of differential calculus, and understanding it thoroughly will open up doors to solving a wide array of problems. We'll also need to recall the Power Rule for differentiating terms like $x^n$ , which states that $d/dx[x^n] = nx^{n-1}$ , and remember that the derivative of a constant is always zero. These foundational calculus basics are the building blocks for applying the Product Rule effectively. Without a solid grasp of these elementary rules, navigating more complex differentiation techniques can be a challenge. So, make sure these fundamental concepts are locked in your brain before we move on to our specific function example. Knowing how to differentiate products isn't just about memorizing the formula, it's about understanding the synergy between individual derivatives.

Deconstructing Our Function: $f(x)=(9x^3+2)(5x^2-4)$

Now that we've refreshed our memory on the Product Rule, let's get down to business with our specific function: $f(x)=(9x^3+2)(5x^2-4)$ . The very first step, guys, is to clearly identify our $u(x)$ and $v(x)$ parts. This might seem super basic, but trust me, getting this wrong at the start can throw off your entire calculation. Take a close look at the function; it's already presented in a perfect product form. So, let's assign them!

We can say:

$u(x) = 9x^3+2$ (This is our first function)
$v(x) = 5x^2-4$ (And this is our second function)

Easy peasy, right? Now comes the fun part: finding their derivatives, $u'(x)$ and $v'(x)$ . We'll use the trusty Power Rule and the rule for differentiating constants (which states that the derivative of any constant is zero). Let's go step-by-step for each one:

Finding $u'(x)$ :

For $u(x) = 9x^3+2$ :

Take the derivative of $9x^3$ : Using the Power Rule, bring the exponent (3) down and multiply it by the coefficient (9), then subtract 1 from the exponent. So, $3 imes 9x^{3-1} = 27x^2$ .
Take the derivative of the constant term, $+2$ : The derivative of any constant is $0$ .

Combine them, and we get: $u'(x) = 27x^2$ .

Finding $v'(x)$ :

Next up, for $v(x) = 5x^2-4$ :

Take the derivative of $5x^2$ : Again, using the Power Rule, bring the exponent (2) down and multiply it by the coefficient (5), then subtract 1 from the exponent. So, $2 imes 5x^{2-1} = 10x^1 = 10x$ .
Take the derivative of the constant term, $-4$ : The derivative of any constant is $0$ .

Combine these, and we have: $v'(x) = 10x$ .

See? It's not so intimidating once you break it down! We've successfully calculated all the individual pieces we need: $u(x)$ , $v(x)$ , $u'(x)$ , and $v'(x)$ . These are the crucial calculus steps before we plug everything into the main Product Rule formula. Showing your work clearly like this isn't just good practice; it helps you catch any potential derivative of polynomial errors and ensures you fully understand the power rule application at each stage. Remember, precision here makes the rest of the problem flow smoothly. Getting these initial derivative calculations correct is paramount for a successful outcome in applying the Product Rule. Don't rush this stage; it's where the foundation of your solution is built. This approach to identifying u and v and then finding their individual derivatives is a cornerstone of effective problem-solving in differential calculus.

Applying the Product Rule: The Grand Calculation!

Alright, calculus crew, this is where all our hard work comes together! We've meticulously identified our functions and their derivatives. Now, it's time to unleash the full power of the Product Rule formula: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . We're going to plug in all the pieces we found. Remember, careful substitution is key here to avoid any silly mistakes. Let's substitute our values:

$u(x) = 9x^3+2$
$v(x) = 5x^2-4$
$u'(x) = 27x^2$
$v'(x) = 10x$

Plugging these into the formula, we get:

$f'(x) = (27x^2)(5x^2-4) + (9x^3+2)(10x)$

This is the core of our derivative solution using the Product Rule calculation. Notice how $u'(x)$ multiplies $v(x)$ (the original second function), and $u(x)$ (the original first function) multiplies $v'(x)$ . The order within each product term doesn't matter (multiplication is commutative), but the overall structure $u'v + uv'$ is important. Now, here's where we need to pay super close attention to the specific format the problem asks for. The problem gives us a template to fill in:

$f^{\prime}(x)=[?] x\left(9 x^3+2\right)+\square x\left(5 x^2-4\right)$

Let's compare our derived expression, $f'(x) = (27x^2)(5x^2-4) + (9x^3+2)(10x)$ , with this template. We need to match the terms to figure out what goes into [?] and \[?] (square).

Looking at the second part of our derivative: $(9x^3+2)(10x)$ . This directly corresponds to the first part of the template: $[?] x(9x^3+2)$ .

See it? We have $(9x^3+2)$ in both. If we rearrange our term $(9x^3+2)(10x)$ to be $(10x)(9x^3+2)$ , it perfectly aligns. So, comparing $(10x)$ with $[?]x$ , it becomes clear that [?] = 10. Boom, one down!

Now, let's look at the first part of our derivative: $(27x^2)(5x^2-4)$ . This needs to correspond to the second part of the template: $\square x(5x^2-4)$ .

Again, we have $(5x^2-4)$ in both. So we need to match $(27x^2)$ with $\square x$ . If $\square x = 27x^2$ , we can solve for $\square$ by dividing both sides by $x$ (assuming $x eq 0$ , which is generally fine for these types of coefficient matching problems). So, $\square = 27x^2 / x = 27x$ . Therefore, \[?] = 27x. Another one solved! This matching coefficients step is a bit tricky, as \[?] isn't a simple number, but an expression involving x. This is an important detail for this specific calculus example. It really tests your algebraic manipulation skills alongside your differentiation knowledge. Understanding how to compare and align terms in different formats is a valuable skill in mathematics.

Simplifying and Understanding the Result

Okay, team, we've found the components that fit into those tricky blanks. But in most calculus problems, you'll want to take it a step further and fully simplify the entire derivative expression. This makes the answer cleaner, easier to analyze, and often reveals underlying patterns. Let's take our derived expression, $f'(x) = (27x^2)(5x^2-4) + (9x^3+2)(10x)$ , and expand it out.

First, let's expand the first term: $(27x^2)(5x^2-4)$

$27x^2 imes 5x^2 = 135x^{2+2} = 135x^4$
$27x^2 imes -4 = -108x^2$

So, the first part becomes: $135x^4 - 108x^2$ .

Next, let's expand the second term: $(9x^3+2)(10x)$

$9x^3 imes 10x = 90x^{3+1} = 90x^4$
$2 imes 10x = 20x$

So, the second part becomes: $90x^4 + 20x$ .

Now, we combine these two expanded parts:

$f'(x) = (135x^4 - 108x^2) + (90x^4 + 20x)$

The final step in our simplifying derivative process is to combine any like terms. Look for terms with the same power of $x$ :

For $x^4$ : $135x^4 + 90x^4 = 225x^4$
For $x^2$ : $-108x^2$ (no other $x^2$ terms)
For $x$ : $20x$ (no other $x$ terms)

Putting it all together, the fully simplified derivative is:

$f'(x) = 225x^4 - 108x^2 + 20x$

This simplified form is usually what's expected for a final answer in most calculus problems. Why is simplification so important? Well, a simplified expression is much easier to work with if you need to do further analysis, like finding critical points, inflection points, or simply evaluating the derivative at a specific $x$ -value. It makes the derivative interpretation much clearer. This process of algebra in calculus is crucial; your algebra skills are just as important as your differentiation skills! It's not just about getting to a correct answer, but about presenting it in its most elegant and usable form. While the problem asked for specific coefficients in a template, understanding how to arrive at this fully expanded and combined polynomial is fundamental. It truly demonstrates a complete grasp of the problem, from initial differentiation to final algebraic simplification, solidifying your understanding of the calculus example we tackled. This comprehensive approach ensures you don't just solve problems, but truly understand them.

Beyond the Basics: When to Use the Product Rule (and When Not To!)

Fantastic job getting through that complex derivative, everyone! Now, let's zoom out a bit and talk about when the Product Rule is your go-to tool, and when you might want to consider other calculus strategies. This is a crucial skill in calculus: knowing which rule to apply and when. It's all about identifying the structure of the function you're working with.

When the Product Rule is a Must:

You absolutely need the Product Rule when you have two functions multiplied together that cannot be easily combined into a simpler form before differentiation. Think about functions like:

$g(x) = e^x \sin(x)$ (Here, you can't multiply $e^x$ and $\sin(x)$ together to simplify them.)
$h(x) = x^2 \ln(x)$ (Again, these are fundamentally different types of functions.)
$k(x) = (x^3+x) \cos(x)$ (Even though $x^3+x$ is a polynomial, multiplying it by $\cos(x)$ creates a term that needs the Product Rule.)

In these cases, trying to differentiate term by term without the Product Rule would lead to an incorrect answer. The individual derivatives of $e^x$ , $\sin(x)$ , $x^2$ , $\ln(x)$ , etc., are known, but their product's derivative requires this specific rule.

When You Can Use It, But Might Not Need It:

Sometimes, you have a product of functions that could be differentiated using the Product Rule, but it might be easier to simplify the function first. Consider $p(x) = (x+1)(x-1)$ .

Using the Product Rule:
- $u = x+1$ , $u' = 1$
- $v = x-1$ , $v' = 1$
- $p'(x) = u'v + uv' = (1)(x-1) + (x+1)(1) = x-1 + x+1 = 2x$
Simplifying First:
- $p(x) = (x+1)(x-1) = x^2 - 1$ (This is a difference of squares!)
- Then, $p'(x) = 2x$ (Using the Power Rule directly)

Both methods give $2x$ , but simplifying first was much quicker! This teaches us a valuable lesson in when to differentiate: always look for opportunities to simplify the function algebraically before you dive into complex differentiation rules. This strategic thinking will save you time and reduce the chances of errors. It's part of developing good advanced differentiation habits. Always ask yourself: