Differentiating Y = (5x³ - 2x + 1) / (x² + 4) A Comprehensive Guide

Hey guys! Let's dive into the exciting world of calculus and tackle a classic differentiation problem. We've got a function here, a fraction involving polynomials, and our mission is to find its derivative. Don't worry, it's not as intimidating as it looks! We'll break it down step by step, making sure everyone understands the process. So, grab your pencils, open your notebooks, and let's get started!

Understanding the Quotient Rule

When you're faced with differentiating a function that's a ratio of two other functions, like our $y = \frac{5x^3 - 2x + 1}{x^2 + 4}$, the quotient rule is your best friend. This rule is a fundamental concept in calculus, specifically designed to handle these kinds of situations. So, what exactly is the quotient rule? Well, it states that if you have a function $y$ that can be expressed as $y = \frac{u}{v}$, where both $u$ and $v$ are themselves functions of $x$, then the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$, can be found using the following formula:

$$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$

Let's break this down even further. In our formula:

• $u$ represents the numerator of the original function.
• $v$ represents the denominator of the original function.
• $\frac{du}{dx}$ is the derivative of the numerator $u$ with respect to $x$.
• $\frac{dv}{dx}$ is the derivative of the denominator $v$ with respect to $x$.
• $v^2$ is simply the square of the denominator $v$.

The quotient rule might look a bit complex at first glance, but it's a powerful tool once you get the hang of it. It's all about systematically finding the derivatives of the numerator and denominator, plugging them into the formula, and then simplifying the result. Trust me, with practice, you'll be wielding the quotient rule like a pro! To really master it, remember that the order of operations in the numerator is crucial – you subtract $u \frac{dv}{dx}$ from $v \frac{du}{dx}$, not the other way around. The denominator is more straightforward; it’s just the square of the original denominator. This rule is a cornerstone of calculus, especially when dealing with rational functions, so understanding it thoroughly will serve you well in many differentiation problems. Keep this formula handy, because we're about to apply it to our function! Remember, practice makes perfect, so the more you use the quotient rule, the more comfortable and confident you'll become with it. We're going to walk through a detailed example next, so you can see the quotient rule in action and understand how to apply it step-by-step.

Identifying u and v

Alright, now that we've got the quotient rule in our toolkit, let's apply it to our specific problem: differentiating $y = \frac{5x^3 - 2x + 1}{x^2 + 4}$. The first crucial step is to correctly identify our $u$ and $v$. Remember, the quotient rule is all about dealing with fractions where both the numerator and the denominator are functions of $x$. In our case, it's pretty clear:

• $u$ is the numerator, which is $5x^3 - 2x + 1$.
• $v$ is the denominator, which is $x^2 + 4$.

So, we've successfully broken down our complex fraction into two simpler functions. This is a key strategy in calculus – often, simplifying a problem into smaller, manageable parts makes the whole process much easier. Once you've identified $u$ and $v$, the next step is to find their derivatives, $\frac{du}{dx}$ and $\frac{dv}{dx}$. This is where our knowledge of basic differentiation rules comes into play. We'll need to use the power rule, which states that the derivative of $x^n$ is $nx^{n-1}$, and remember that the derivative of a constant is zero. By carefully applying these rules, we can find the derivatives of both $u$ and $v$. It’s important to be meticulous in this step, because any errors in finding $\frac{du}{dx}$ or $\frac{dv}{dx}$ will propagate through the rest of the problem. Double-checking your work at this stage can save you a lot of headaches later on! So, we've got $u$ and $v$ identified, and we're ready to move on to the next stage: finding their derivatives. This is where things start to get really interesting, as we'll be putting our differentiation skills to the test. Just remember to take your time, apply the rules carefully, and you'll be well on your way to solving this problem. We're building a solid foundation step-by-step, and each step brings us closer to the final answer. Next, we'll calculate $\frac{du}{dx}$ and $\frac{dv}{dx}$, and then we can plug everything into the quotient rule formula.

Calculating du/dx and dv/dx

Okay, we've identified our $u$ and $v$. Now comes the fun part: finding their derivatives. This is where we put our basic differentiation skills to the test. Remember, we have:

• $u = 5x^3 - 2x + 1$
• $v = x^2 + 4$

Let's start with finding $\frac{du}{dx}$, the derivative of $u$ with respect to $x$. We'll apply the power rule term by term. The power rule, if you recall, states that the derivative of $x^n$ is $nx^{n-1}$. So, for the first term, $5x^3$, we multiply the coefficient (5) by the exponent (3) and reduce the exponent by 1, giving us $15x^2$. For the second term, $-2x$, the derivative is simply -2 (remember, $x$ is the same as $x^1$, so the derivative is $1 \cdot x^0 = 1$). And finally, the derivative of the constant term, 1, is zero. Putting it all together, we get:

$$\frac{du}{dx} = 15x^2 - 2$$

Now, let's find $\frac{dv}{dx}$, the derivative of $v$ with respect to $x$. We'll use the same power rule approach. For the first term, $x^2$, the derivative is $2x$. For the second term, 4, which is a constant, the derivative is zero. So, we have:

$$\frac{dv}{dx} = 2x$$

And there you have it! We've successfully calculated the derivatives of both $u$ and $v$. This is a critical step in applying the quotient rule, and it's essential to get these derivatives right. A small mistake here can throw off the entire solution, so it's always a good idea to double-check your work. Now that we have $\frac{du}{dx}$ and $\frac{dv}{dx}$, we're ready to plug them into the quotient rule formula. This is where the pieces of the puzzle start to come together. We'll substitute our expressions for $u$, $v$, $\frac{du}{dx}$, and $\frac{dv}{dx}$ into the formula, and then we'll simplify the resulting expression. This simplification process might involve some algebraic manipulation, but don't worry, we'll take it step by step. We're making great progress, guys! We've tackled the tricky part of finding the derivatives, and now we're ready to assemble the final answer. So, let's move on to the next step: plugging everything into the quotient rule formula and simplifying.

Applying the Quotient Rule Formula

Awesome! We've done the groundwork – identifying $u$ and $v$, and calculating their derivatives. Now, the moment we've been waiting for: applying the quotient rule formula. Let's remind ourselves of the formula:

$$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$

We know:

• $u = 5x^3 - 2x + 1$
• $v = x^2 + 4$
• $\frac{du}{dx} = 15x^2 - 2$
• $\frac{dv}{dx} = 2x$

Now, let's substitute these values into the formula. This is where careful attention to detail is crucial. We need to make sure we're plugging everything in the right place and that we're using parentheses correctly to maintain the proper order of operations. So, let's do it:

$$\frac{dy}{dx} = \frac{(x^2 + 4)(15x^2 - 2) - (5x^3 - 2x + 1)(2x)}{(x^2 + 4)^2}$$

Okay, we've got a fairly complex expression now. But don't be intimidated! The next step is to simplify this. This will involve expanding the products in the numerator and then combining like terms. Remember, the goal is to get the expression into its simplest form. This is where your algebra skills will shine! We'll need to carefully distribute the terms and then combine any terms that have the same power of $x$. This might seem like a lot of work, but it's a necessary step to get to the final answer. The denominator, $(x^2 + 4)^2$, can either be left as it is or expanded, depending on the context and what the problem asks for. Sometimes, leaving the denominator in factored form can be more useful, especially if you need to analyze the behavior of the derivative. So, we've plugged everything into the quotient rule formula, and we're ready for the next challenge: simplifying the expression. This is where we'll really see the power of the quotient rule come to fruition, as we transform this complex expression into a more manageable form. Let's roll up our sleeves and get ready to simplify!

Simplifying the Expression

Alright, we've got the derivative expressed using the quotient rule, but it looks a bit… messy. The next crucial step is to simplify the expression. This involves expanding the products in the numerator and then combining like terms. It's like tidying up after a math party – we need to organize everything and make it look neat and presentable. Let's start by expanding the numerator:

$$\frac{dy}{dx} = \frac{(x^2 + 4)(15x^2 - 2) - (5x^3 - 2x + 1)(2x)}{(x^2 + 4)^2}$$

First, we'll expand $(x^2 + 4)(15x^2 - 2)$. Using the distributive property (or the FOIL method), we get:

$$15x^4 - 2x^2 + 60x^2 - 8 = 15x^4 + 58x^2 - 8$$

Next, we'll expand $(5x^3 - 2x + 1)(2x)$:

$$10x^4 - 4x^2 + 2x$$

Now, let's substitute these expanded expressions back into the numerator:

$$\frac{dy}{dx} = \frac{(15x^4 + 58x^2 - 8) - (10x^4 - 4x^2 + 2x)}{(x^2 + 4)^2}$$

Remember to distribute the negative sign in the second term:

$$\frac{dy}{dx} = \frac{15x^4 + 58x^2 - 8 - 10x^4 + 4x^2 - 2x}{(x^2 + 4)^2}$$

Now, we combine like terms in the numerator:

$$\frac{dy}{dx} = \frac{5x^4 + 62x^2 - 2x - 8}{(x^2 + 4)^2}$$

And there you have it! We've successfully simplified the numerator. The denominator, $(x^2 + 4)^2$, can be left as is, or you can expand it if you need to for a particular application. However, in many cases, leaving it in factored form is perfectly acceptable. This simplified expression is the derivative of our original function. It might have looked daunting at first, but by breaking it down step by step and carefully applying the quotient rule and algebraic simplification techniques, we've arrived at the solution. This process highlights the power of calculus and the importance of mastering fundamental algebraic skills. So, we've simplified the expression and found the derivative! We took a complex-looking fraction and transformed it into a more manageable form. This is a testament to the power of calculus and algebraic manipulation. But remember, the journey doesn't end here. It’s essential to practice more problems to solidify your understanding and build confidence in your skills.

The Final Answer

Drumroll, please! After all our hard work, we've arrived at the final answer. We started with a function, $y = \frac{5x^3 - 2x + 1}{x^2 + 4}$, and we wanted to find its derivative, $\frac{dy}{dx}$. We've successfully navigated the quotient rule, calculated derivatives, and simplified the expression. So, what's the grand finale? Here it is:

$$\frac{dy}{dx} = \frac{5x^4 + 62x^2 - 2x - 8}{(x^2 + 4)^2}$$

This is the derivative of our original function. It tells us the instantaneous rate of change of $y$ with respect to $x$ at any point. In other words, it's the slope of the tangent line to the curve of the function at any given $x$ value. That's pretty cool, right? This answer represents the culmination of all our efforts. We've applied the quotient rule, a fundamental concept in calculus, and we've used our algebraic skills to simplify the expression. This is a typical example of how calculus problems are solved – it's a combination of applying rules and techniques and then using algebra to massage the expression into its final form. But more than just an answer, this represents a journey of problem-solving. We encountered a challenge, we broke it down into smaller parts, we applied the appropriate tools, and we arrived at a solution. This is a valuable skill that extends far beyond the realm of mathematics. Whether you're solving a complex equation or tackling a real-world problem, the ability to break it down, apply the right techniques, and persevere until you find a solution is essential. So, celebrate this victory! You've successfully differentiated a complex function using the quotient rule. And remember, the more you practice, the more confident and skilled you'll become. Keep exploring the world of calculus, and you'll discover even more fascinating and powerful tools for solving problems. Congratulations, guys! We did it!