Derivative Of F(x) = (5x^2 + 4) / (3 + X^2)

Hey guys, today we're diving into the fascinating world of calculus to tackle a derivative problem. We've got a function, f(x)=rac{5 x^2+4}{3+x^2}, and our mission is to find its derivative, $f^{\prime}(x)$, in its most simplified form. This might sound a bit intimidating, but don't worry, we'll break it down step-by-step. Understanding derivatives is super important in mathematics, especially when you're dealing with rates of change, optimization problems, and understanding the behavior of functions. So, grab your thinking caps, and let's get started on unraveling this calculus mystery!

Understanding the Quotient Rule

Before we jump into finding the derivative of our specific function, f(x)=rac{5 x^2+4}{3+x^2}, it's crucial to have a solid grasp of the quotient rule. This rule is our go-to tool when we need to differentiate a function that's expressed as a fraction, where one function is divided by another. Think of it like this: if you have a function $f(x)$ that can be written as $\frac{u(x)}{v(x)}$, the quotient rule gives us a straightforward way to find its derivative, $f^{\prime}(x)$. The formula for the quotient rule is:

$$f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{[v(x)]^2}$$

Let's break down what this formula means. First, we need to identify our 'u(x)' and 'v(x)'. In our case, f(x)=rac{5 x^2+4}{3+x^2}, so we can set:

• $u(x) = 5x^2 + 4$ (the numerator)
• $v(x) = 3 + x^2$ (the denominator)

Next, we need to find the derivatives of these two parts: $u^{\prime}(x)$ and $v^{\prime}(x)$. This is where the power rule comes in handy, which states that the derivative of $ax^n$ is $anx^{n-1}$.

For $u(x) = 5x^2 + 4$, its derivative $u^{\prime}(x)$ is found by taking the derivative of each term:

• The derivative of $5x^2$ is $5 \times 2x^{2-1} = 10x^1 = 10x$.
• The derivative of a constant (like 4) is always 0.

So, $u^{\prime}(x) = 10x + 0 = 10x$.

For $v(x) = 3 + x^2$, its derivative $v^{\prime}(x)$ is found similarly:

• The derivative of a constant (like 3) is 0.
• The derivative of $x^2$ is $2x^{2-1} = 2x^1 = 2x$.

So, $v^{\prime}(x) = 0 + 2x = 2x$.

Now that we have $u(x)$, $v(x)$, $u^{\prime}(x)$, and $v^{\prime}(x)$, we can plug them all into the quotient rule formula. This is where the real work begins, and paying attention to detail is key to avoiding silly mistakes. We're going to substitute these pieces into the quotient rule structure and then focus on simplifying the resulting expression.

Applying the Quotient Rule to Our Function

Alright, we've identified our components and their derivatives. Now, let's plug $u(x) = 5x^2 + 4$, $v(x) = 3 + x^2$, $u^{\prime}(x) = 10x$, and $v^{\prime}(x) = 2x$ into the quotient rule formula: $f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{[v(x)]^2}$.

$$f^{\prime}(x) = \frac{(10x)(3 + x^2) - (5x^2 + 4)(2x)}{(3 + x^2)^2}$$

This is our derivative, but it's not in its simplified form yet. The next crucial step is to algebraically simplify the numerator. This usually involves expanding terms, combining like terms, and generally making the expression as neat as possible. Let's tackle the numerator first:

Expand the first term: $(10x)(3 + x^2) = 10x \times 3 + 10x \times x^2 = 30x + 10x^3$.

Expand the second term: $(5x^2 + 4)(2x) = 5x^2 \times 2x + 4 \times 2x = 10x^3 + 8x$.

Now, substitute these expanded forms back into the numerator, remembering the minus sign in front of the second term:

Numerator = $(30x + 10x^3) - (10x^3 + 8x)$

When we distribute the negative sign, it becomes:

Numerator = $30x + 10x^3 - 10x^3 - 8x$

Now, let's combine the like terms in the numerator:

• The $10x^3$ and $-10x^3$ terms cancel each other out.
• We are left with $30x - 8x = 22x$.

So, the simplified numerator is $22x$.

Now, we can rewrite our derivative with the simplified numerator:

$$f^{\prime}(x) = \frac{22x}{(3 + x^2)^2}$$

And there you have it! We've successfully applied the quotient rule and simplified the expression. The denominator, $(3 + x^2)^2$, is usually left in its factored form unless there's a specific reason to expand it, as expanding it often makes the expression more complex rather than simpler. This final form is clean, concise, and represents the derivative of our original function. Keep practicing these steps, guys, and you'll become a derivative pro in no time!

The Importance of Simplified Form

So, why do we stress about getting our derivative into a simplified form? It’s not just about making our answers look pretty; simplification plays a really important role in mathematics and has practical implications. When we simplify an expression, like we did with $f^{\prime}(x)$, we make it easier to understand, analyze, and use in further calculations. Imagine trying to solve an equation with a super complicated expression versus a simple one – the simple one is always going to be easier to work with, right?

In the context of derivatives, a simplified form allows us to more easily:

1. Find critical points: Critical points occur where the derivative is either zero or undefined. A simplified derivative makes it much faster to set the numerator to zero and solve for $x$. In our case, setting $22x = 0$ immediately tells us that $x=0$ is a critical point. This is way easier than trying to solve an equation derived from an unsimplified numerator.
2. Determine intervals of increase and decrease: By looking at the sign of the derivative in different intervals, we can tell if the original function is increasing or decreasing. A simplified form of $f^{\prime}(x)$ makes it easier to test these intervals. For $f^{\prime}(x) = \frac{22x}{(3 + x^2)^2}$, since the denominator $(3 + x^2)^2$ is always positive (because anything squared is non-negative, and $3+x^2$ is never zero), the sign of $f^{\prime}(x)$ is determined solely by the sign of the numerator, $22x$. This means $f^{\prime}(x) > 0$ when $x > 0$ (function is increasing) and $f^{\prime}(x) < 0$ when $x < 0$ (function is decreasing).
3. Find the concavity and inflection points: While the first derivative tells us about the slope, the second derivative tells us about the curvature. To find the second derivative, we'd need to differentiate $f^{\prime}(x)$ again, and having $f^{\prime}(x)$ in a simplified form is absolutely essential for that process.
4. Understand the behavior of the function: A simplified derivative gives us a clearer picture of how the original function is behaving. We can quickly see where the function is steep, flat, increasing, or decreasing.
5. Communicate mathematical ideas: When you present your work, a simplified answer is generally expected and appreciated. It shows that you've gone the extra mile to present your solution clearly and efficiently.

So, even though it might take a few extra minutes of algebraic manipulation, simplifying your derivative is a crucial step that makes all subsequent analysis much smoother and more accurate. It's like cleaning up your workspace before starting a big project – it makes everything else easier!

Conclusion

We've successfully navigated the process of finding the derivative of f(x)=rac{5 x^2+4}{3+x^2} using the quotient rule. We identified the numerator $u(x)=5x^2+4$ and the denominator $v(x)=3+x^2$, found their respective derivatives $u^{\prime}(x)=10x$ and $v^{\prime}(x)=2x$, and applied the quotient rule formula: $f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{[v(x)]^2}$.

After careful substitution and algebraic simplification, particularly in the numerator, we arrived at the final, simplified derivative:

$$f^{\prime}(x) = \frac{22x}{(3 + x^2)^2}$$

This result highlights the power of calculus rules and algebraic manipulation. Remember, guys, the key takeaways are to master the quotient rule and to always strive for simplification. A simplified derivative isn't just an aesthetic choice; it's a practical necessity for further analysis, understanding function behavior, and making complex problems more manageable. Keep practicing, stay curious, and you'll conquer any derivative challenge that comes your way!