Finding Dy/dx For Y=(x^2+1)(3x-2) A Step-by-Step Solution

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In the realm of calculus, the derivative of a function serves as a cornerstone, offering invaluable insights into the function's rate of change. This article embarks on a comprehensive exploration of the derivative of the function y = (x^2 + 1)(3x - 2), a polynomial expression that gracefully intertwines quadratic and linear components. Our journey will commence with a meticulous dissection of the problem, unraveling the intricacies of differentiation rules that govern the manipulation of such expressions. We will then embark on a step-by-step odyssey, meticulously applying these rules to arrive at the derivative, dy/dx. Finally, we will delve into the profound implications of this derivative, illuminating its significance in discerning the function's behavior and its far-reaching applications in diverse fields.

Before we delve into the intricacies of finding the derivative, let's first gain a comprehensive understanding of the problem at hand. We are presented with the function y = (x^2 + 1)(3x - 2), a polynomial expression elegantly crafted by the product of a quadratic term (x^2 + 1) and a linear term (3x - 2). Our mission is to determine the derivative of this function, denoted as dy/dx, which represents the instantaneous rate of change of y with respect to x. In essence, the derivative unveils how the function y responds to minute variations in the independent variable x. This concept forms the bedrock of calculus, enabling us to analyze the function's behavior, identify critical points, and gain profound insights into its overall characteristics. To unravel the derivative of this function, we will judiciously employ the product rule, a fundamental principle in calculus that governs the differentiation of products of functions. This rule will serve as our guiding light, leading us through the intricate steps of differentiation and ultimately revealing the derivative we seek.

At the heart of calculus lies a set of fundamental rules that govern the differentiation of various types of functions. These rules serve as the bedrock upon which we construct our understanding of derivatives and their applications. Let's take a moment to review some of these essential rules, which will prove instrumental in our quest to find the derivative of y = (x^2 + 1)(3x - 2).

  • The Power Rule: This rule reigns supreme when dealing with power functions, those elegant expressions of the form x^n, where n is a constant. The power rule dictates that the derivative of x^n is simply n times x raised to the power of n - 1. In mathematical notation, this is expressed as d/dx (x^n) = nx^(n-1). This rule will be our steadfast companion as we navigate the differentiation process.
  • The Constant Multiple Rule: This rule gracefully handles constants that accompany functions. It states that the derivative of a constant c multiplied by a function f(x) is simply the constant c times the derivative of the function f(x). In mathematical terms, this is represented as d/dx [c * f(x)] = c * d/dx [f(x)]. This rule allows us to seamlessly integrate constants into our differentiation calculations.
  • The Sum/Difference Rule: This rule elegantly governs the differentiation of sums and differences of functions. It states that the derivative of a sum or difference of functions is simply the sum or difference of their individual derivatives. Mathematically, this is expressed as d/dx [f(x) ± g(x)] = d/dx [f(x)] ± d/dx [g(x)]. This rule empowers us to break down complex functions into simpler components and differentiate them individually.
  • The Product Rule: This rule, the cornerstone of our endeavor, dictates how to differentiate the product of two functions. It states that the derivative of the product of two functions, u(x) and v(x), is given by u'(x)v(x) + u(x)v'(x), where u'(x) and v'(x) represent the derivatives of u(x) and v(x), respectively. In mathematical notation, this is expressed as d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). This rule will be our guiding light as we unravel the derivative of y = (x^2 + 1)(3x - 2).

With these fundamental rules firmly in our grasp, we are now fully equipped to embark on the journey of finding the derivative of our target function.

Now, let's embark on a step-by-step journey to unravel the derivative of the function y = (x^2 + 1)(3x - 2). Our primary tool in this endeavor will be the product rule, a fundamental principle in calculus that governs the differentiation of products of functions.

Step 1: Identifying u(x) and v(x)

The first step in applying the product rule is to identify the two functions, u(x) and v(x), that are being multiplied together. In our case, we can readily identify them as follows:

  • u(x) = x^2 + 1
  • v(x) = 3x - 2

These two functions, a quadratic expression and a linear expression, respectively, form the building blocks of our target function.

Step 2: Finding u'(x) and v'(x)

Next, we need to determine the derivatives of both u(x) and v(x). This involves applying the power rule and the constant multiple rule, which we discussed earlier.

  • For u(x) = x^2 + 1, the derivative u'(x) is found as follows:
    • d/dx (x^2) = 2x (Applying the power rule)
    • d/dx (1) = 0 (The derivative of a constant is zero)
    • Therefore, u'(x) = 2x + 0 = 2x
  • For v(x) = 3x - 2, the derivative v'(x) is found as follows:
    • d/dx (3x) = 3 (Applying the constant multiple rule and the power rule)
    • d/dx (-2) = 0 (The derivative of a constant is zero)
    • Therefore, v'(x) = 3 - 0 = 3

We have now successfully determined the derivatives of both u(x) and v(x), which are essential components for applying the product rule.

Step 3: Applying the Product Rule

With u(x), v(x), u'(x), and v'(x) in hand, we are now ready to invoke the product rule. Recall that the product rule states:

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

Substituting the expressions we found earlier, we get:

dy/dx = (2x)(3x - 2) + (x^2 + 1)(3)

This expression represents the derivative of our target function, but we can simplify it further to obtain a more elegant form.

Step 4: Simplifying the Expression

To simplify the expression, we will expand the products and combine like terms:

dy/dx = 6x^2 - 4x + 3x^2 + 3

Combining the x^2 terms, we arrive at:

dy/dx = 9x^2 - 4x + 3

This is the final, simplified expression for the derivative of y = (x^2 + 1)(3x - 2).

The derivative we have diligently obtained, dy/dx = 9x^2 - 4x + 3, is not merely a mathematical expression; it is a powerful lens through which we can glean profound insights into the behavior of the original function, y = (x^2 + 1)(3x - 2). Let's delve into the implications of this derivative and explore its significance.

  • Rate of Change: The derivative, at its core, represents the instantaneous rate of change of the function y with respect to x. In simpler terms, it tells us how much the function's value changes for a minuscule change in the input x. For instance, if we evaluate the derivative at a specific point x = a, the resulting value, dy/dx|_(x=a), signifies the slope of the tangent line to the graph of y at that point. A positive derivative indicates that the function is increasing at that point, while a negative derivative signifies a decreasing function. A derivative of zero implies a critical point, where the function may reach a local maximum, local minimum, or a saddle point.
  • Critical Points: Critical points, those pivotal locations where the derivative equals zero or is undefined, hold immense significance in understanding a function's behavior. At these points, the function's slope momentarily vanishes, often indicating a transition from increasing to decreasing or vice versa. To identify critical points, we set the derivative equal to zero and solve for x. In our case, we need to solve the quadratic equation 9x^2 - 4x + 3 = 0. The solutions to this equation will reveal the critical points of the function y.
  • Local Maxima and Minima: Critical points often mark the locations of local maxima and minima, those peaks and valleys in the function's graph. To determine whether a critical point corresponds to a local maximum, local minimum, or neither, we can employ the second derivative test. This test involves evaluating the second derivative of the function at the critical point. A negative second derivative indicates a local maximum, a positive second derivative suggests a local minimum, and a second derivative of zero implies that the test is inconclusive.
  • Applications in Optimization: The derivative's ability to pinpoint critical points makes it an indispensable tool in optimization problems. These problems often involve finding the maximum or minimum value of a function subject to certain constraints. The derivative helps us identify the potential locations of these extreme values, guiding us towards the optimal solution. Optimization problems arise in a multitude of fields, from engineering design to economic modeling.

The derivative, a concept born from the realm of mathematics, transcends its theoretical origins and finds practical applications in a diverse array of real-world scenarios. Its ability to quantify rates of change and identify critical points makes it an invaluable tool in various disciplines. Let's explore some of these fascinating applications.

  • Physics: In the realm of physics, the derivative reigns supreme as the quantifier of motion. Velocity, the rate of change of position with respect to time, is simply the first derivative of the position function. Acceleration, the rate of change of velocity with respect to time, is the second derivative of the position function. These derivatives allow physicists to meticulously analyze the motion of objects, from the trajectory of a projectile to the oscillations of a pendulum. The derivative also plays a pivotal role in understanding other physical phenomena, such as heat transfer, wave propagation, and the dynamics of electric circuits.
  • Engineering: Engineers, the architects of our technological world, rely heavily on derivatives to design and optimize structures, machines, and systems. In structural engineering, derivatives are used to calculate stresses and strains in materials, ensuring the stability and safety of bridges, buildings, and other structures. In mechanical engineering, derivatives are employed to analyze the motion of machines, optimize engine performance, and design control systems. Electrical engineers use derivatives to model and analyze circuits, design filters, and optimize signal processing algorithms.
  • Economics: Economics, the science of resource allocation, harnesses the power of derivatives to model and analyze economic systems. Marginal cost, the rate of change of cost with respect to production, is a derivative that helps businesses determine the optimal production level. Marginal revenue, the rate of change of revenue with respect to sales, is another derivative that guides pricing and marketing decisions. Derivatives also play a crucial role in financial modeling, where they are used to price derivatives, manage risk, and forecast market trends.
  • Computer Science: In the realm of computer science, derivatives find applications in machine learning, optimization algorithms, and computer graphics. Gradient descent, a fundamental algorithm in machine learning, relies on derivatives to adjust model parameters and minimize errors. Derivatives are also used in optimization algorithms to find the optimal solutions to various problems, such as resource allocation and scheduling. In computer graphics, derivatives are employed to create smooth curves and surfaces, simulate realistic lighting effects, and generate animations.

In this comprehensive exploration, we embarked on a journey to unveil the derivative of the function y = (x^2 + 1)(3x - 2). We meticulously applied the product rule, a cornerstone of calculus, to arrive at the derivative dy/dx = 9x^2 - 4x + 3. This derivative, however, is not merely a mathematical expression; it is a window into the function's behavior, revealing its rate of change, critical points, and potential for optimization.

We delved into the implications of the derivative, understanding its role in determining the function's increasing and decreasing intervals, identifying local maxima and minima, and solving optimization problems. Furthermore, we transcended the theoretical realm and explored the derivative's diverse applications in real-world scenarios, from physics and engineering to economics and computer science.

The derivative, a concept that may seem abstract at first glance, emerges as a powerful tool for understanding and manipulating the world around us. Its ability to quantify change and identify critical points makes it an indispensable asset in various disciplines, driving innovation and progress.

As we conclude this exploration, let us carry forward the understanding and appreciation of the derivative, recognizing its profound significance in the tapestry of mathematics and its far-reaching impact on our world.

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