Derivative Of H(x) = (x^2 + 1) / (2x^2 - 3x) - Calculus Guide
Calculus can be challenging, especially when dealing with derivatives of complex functions. In this comprehensive guide, we will break down the process of finding the derivative of the function h(x) = (x^2 + 1) / (2x^2 - 3x). We'll walk through each step with clear explanations and examples to ensure you grasp the concept thoroughly. Whether you're a student tackling calculus problems or just looking to brush up on your skills, this article will provide you with the knowledge and confidence you need. So, let's dive in and explore the world of derivatives!
Understanding the Quotient Rule
Before we dive into the specifics of our function, it's crucial to understand the quotient rule. The quotient rule is a fundamental concept in calculus that allows us to find the derivative of a function that is expressed as the quotient of two other functions. Essentially, it provides a formula for differentiating fractions where both the numerator and the denominator are functions of x. This rule is particularly useful when dealing with rational functions, like the one we're about to analyze. Mastering the quotient rule is essential for anyone studying calculus, as it appears in numerous applications and problems. Let's delve deeper into the specifics of the quotient rule to ensure we're well-prepared for our example. The quotient rule states that if you have a function h(x) that is defined as the ratio of two other functions, say f(x) and g(x), such that h(x) = f(x) / g(x), then the derivative of h(x), denoted as h'(x), can be found using the following formula:
h'(x) = [f'(x) * g(x) - f(x) * g'(x)] / [g(x)]^2
Where:
- f'(x) is the derivative of the numerator function f(x).
- g'(x) is the derivative of the denominator function g(x).
This formula might seem complex at first glance, but it becomes quite manageable with practice. The key is to correctly identify the numerator and denominator functions, find their respective derivatives, and then plug them into the formula. The denominator of the derivative, [g(x)]^2, is simply the square of the original denominator function. To further illustrate the importance of the quotient rule, consider scenarios where you need to find the rate of change of a ratio, such as the rate of change of population density (population divided by area) or the rate of change of average cost (total cost divided by the number of units produced). In these cases, the quotient rule becomes indispensable. Understanding the quotient rule not only helps in solving mathematical problems but also provides a foundation for more advanced calculus concepts. Now that we have a solid understanding of the quotient rule, let's apply it to our specific function h(x) = (x^2 + 1) / (2x^2 - 3x) and find its derivative step by step.
Applying the Quotient Rule to h(x)
Now, let's apply the quotient rule to find the derivative of our function, h(x) = (x^2 + 1) / (2x^2 - 3x). This involves identifying the numerator and denominator, finding their derivatives, and then plugging these values into the quotient rule formula. This is where our understanding of the rule will be put to the test, and we'll see how it simplifies the process of finding the derivative of a seemingly complex rational function. We'll break it down step by step to make sure we don't miss any crucial details. First, we need to identify our f(x) and g(x). In our case:
- f(x) = x^2 + 1 (the numerator)
- g(x) = 2x^2 - 3x (the denominator)
Next, we need to find the derivatives of f(x) and g(x). These are relatively straightforward applications of the power rule, which states that the derivative of x^n is nx^(n-1). Let's calculate them:
- f'(x) = d/dx (x^2 + 1) = 2x (The derivative of x^2 is 2x, and the derivative of the constant 1 is 0)
- g'(x) = d/dx (2x^2 - 3x) = 4x - 3 (The derivative of 2x^2 is 4x, and the derivative of -3x is -3)
Now that we have f(x), g(x), f'(x), and g'(x), we can plug them into the quotient rule formula:
h'(x) = [f'(x) * g(x) - f(x) * g'(x)] / [g(x)]^2 h'(x) = [(2x) * (2x^2 - 3x) - (x^2 + 1) * (4x - 3)] / (2x^2 - 3x)^2
This is the direct application of the quotient rule formula. The next step involves simplifying the expression, which is where algebraic skills come into play. It's crucial to expand the terms in the numerator and then combine like terms to arrive at a simplified form of the derivative. This simplification process not only makes the derivative easier to work with but also helps in understanding the behavior of the function h(x). Let's proceed with simplifying the numerator to see what we get.
Simplifying the Expression
After applying the quotient rule, we arrived at a somewhat complex expression for h'(x). Now, the crucial step is to simplify this expression. Simplification not only makes the derivative more manageable but also reveals its underlying structure, which can be essential for further analysis or applications. This involves expanding the terms in the numerator, combining like terms, and potentially factoring to arrive at the simplest form of the derivative. Let's dive into the algebraic manipulation required to simplify the expression. We have:
h'(x) = [(2x) * (2x^2 - 3x) - (x^2 + 1) * (4x - 3)] / (2x^2 - 3x)^2
First, let's expand the terms in the numerator:
(2x) * (2x^2 - 3x) = 4x^3 - 6x^2 (x^2 + 1) * (4x - 3) = 4x^3 - 3x^2 + 4x - 3
Now, substitute these expansions back into the numerator:
4x^3 - 6x^2 - (4x^3 - 3x^2 + 4x - 3)
Distribute the negative sign:
4x^3 - 6x^2 - 4x^3 + 3x^2 - 4x + 3
Combine like terms:
(4x^3 - 4x^3) + (-6x^2 + 3x^2) - 4x + 3 = -3x^2 - 4x + 3
So, the simplified numerator is -3x^2 - 4x + 3. Now, we can write the simplified derivative:
h'(x) = (-3x^2 - 4x + 3) / (2x^2 - 3x)^2
This is the simplified form of the derivative of h(x). It's much cleaner and easier to work with than the initial expression we obtained after applying the quotient rule. The denominator, (2x^2 - 3x)^2, remains as is since expanding it would not lead to further significant simplification in this case. This simplified form of h'(x) allows us to analyze the behavior of the function h(x) more effectively. For instance, we can now easily find critical points by setting the numerator equal to zero and solving for x. Understanding the critical points helps us determine where the function is increasing or decreasing. The simplified derivative also makes it easier to compute the slope of the tangent line at any point on the graph of h(x). This is a key application of derivatives in calculus. In the next section, we will discuss some applications and implications of this derivative.
Applications and Implications of the Derivative
Now that we've successfully found and simplified the derivative of h(x), it's essential to understand the applications and implications of this result. Derivatives are not just abstract mathematical constructs; they have practical uses in various fields, including physics, engineering, economics, and computer science. Understanding how to interpret and apply a derivative can provide valuable insights into the behavior of the original function and the system it models. Let's explore some key applications and what our derivative, h'(x) = (-3x^2 - 4x + 3) / (2x^2 - 3x)^2, tells us about the function h(x). One of the primary applications of the derivative is finding the slope of the tangent line at any point on the graph of the function. The value of h'(x) at a specific x-value gives the slope of the line that touches the curve of h(x) at that point. This is incredibly useful for understanding the instantaneous rate of change of the function. For instance, if we want to find the slope of the tangent line at x = 1, we would simply plug x = 1 into h'(x):
h'(1) = (-3(1)^2 - 4(1) + 3) / (2(1)^2 - 3(1))^2 h'(1) = (-3 - 4 + 3) / (2 - 3)^2 h'(1) = -4 / 1 h'(1) = -4
So, the slope of the tangent line to the graph of h(x) at x = 1 is -4. This tells us that the function is decreasing at that point. Another crucial application of derivatives is finding critical points. Critical points are the points where the derivative is either zero or undefined. These points are significant because they can indicate local maxima, local minima, or saddle points of the function. To find the critical points, we need to set the numerator of h'(x) equal to zero and solve for x:
-3x^2 - 4x + 3 = 0
This is a quadratic equation, which we can solve using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
Where a = -3, b = -4, and c = 3. Plugging these values in:
x = [4 ± √((-4)^2 - 4(-3)(3))] / (2(-3)) x = [4 ± √(16 + 36)] / (-6) x = [4 ± √52] / (-6) x = [4 ± 2√13] / (-6) x = [-2 ± √13] / 3
So, the critical points occur at x = (-2 + √13) / 3 and x = (-2 - √13) / 3. We also need to consider where the derivative is undefined, which occurs when the denominator is zero:
(2x^2 - 3x)^2 = 0 2x^2 - 3x = 0 x(2x - 3) = 0 x = 0 or x = 3/2
Thus, the derivative is undefined at x = 0 and x = 3/2. These points, along with the roots of the numerator, are critical points of h(x). By analyzing the sign of h'(x) in the intervals determined by these critical points, we can determine where the function h(x) is increasing or decreasing. Additionally, the second derivative, which we haven't calculated here, can provide information about the concavity of the function. In conclusion, finding the derivative h'(x) is just the first step. The real power comes from understanding how to interpret and apply this derivative to gain insights into the behavior of the original function. From finding tangent lines to identifying critical points and understanding increasing/decreasing intervals, the derivative is a fundamental tool in calculus and its applications. For further exploration of derivatives and their applications, consider visiting Khan Academy's Calculus section, a trusted resource for learning mathematics.
