Finding Dy/dt Using The Chain Rule A Step-by-Step Calculus Solution

This article delves into the process of finding the derivative of y with respect to t, denoted as dy/dt, given a set of interconnected equations. We are presented with a scenario where y is a function of u, u is a function of s, and s is a function of t. This intricate relationship necessitates the application of the chain rule, a fundamental concept in calculus that allows us to differentiate composite functions. Our main keywords include: dy/dt, chain rule, derivatives, composite functions, and calculus. By meticulously applying the chain rule and employing basic differentiation techniques, we will systematically unravel the relationship between y and t, ultimately arriving at the desired derivative, dy/dt. This exploration will not only demonstrate the power of the chain rule but also highlight the importance of understanding how variables interact within a system of equations. To fully grasp the solution, we will break down each step, starting with identifying the individual derivatives and then combining them using the chain rule formula. We will also emphasize the significance of expressing the final answer in terms of the original variable, t, to provide a complete and meaningful solution. The problem at hand exemplifies a classic application of the chain rule, showcasing its versatility in handling complex dependencies between variables. Understanding the chain rule is crucial for solving various problems in calculus and related fields, making this exercise a valuable learning experience. In the subsequent sections, we will walk through the step-by-step solution, providing clear explanations and highlighting key concepts along the way. This will ensure a thorough understanding of the process and empower readers to tackle similar problems with confidence.

We are given the following equations:

• y = 3u + 1
• u = 1/s²
• s = 1 - t

The objective is to find dy/dt. This task involves differentiating a composite function, where y depends on u, u depends on s, and s depends on t. To achieve this, we will leverage the chain rule of calculus. The chain rule states that if y is a function of u, and u is a function of x, then dy/dx = (dy/du) * (du/dx). In our case, we have three functions, so the chain rule will be extended to include the derivative of s with respect to t. This means we need to find dy/du, du/ds, and ds/dt, and then multiply them together to get dy/dt. The challenge lies in correctly identifying each derivative and performing the necessary algebraic manipulations to express the final answer in terms of t. This problem underscores the importance of understanding functional dependencies and the power of the chain rule in navigating these relationships. Moreover, it reinforces the concept of derivatives as rates of change, allowing us to see how a change in t ultimately affects the value of y. The process of solving this problem will not only provide the answer but also deepen our understanding of the chain rule and its applications in calculus. We will break down the problem into smaller, manageable steps, making it easier to follow and comprehend. Each step will be explained in detail, ensuring that the logic behind the solution is clear and transparent.

To find dy/dt, we will use the chain rule. Since y is a function of u, u is a function of s, and s is a function of t, we can express dy/dt as follows:

dy/dt = (dy/du) * (du/ds) * (ds/dt)

Let's find each derivative individually:

1. dy/du: Given y = 3u + 1, we differentiate y with respect to u: dy/du = 3

   This derivative is straightforward since the equation is linear. The derivative of a constant is zero, and the derivative of 3u with respect to u is simply 3. This result tells us that for every unit change in u, y changes by 3 units. This is a crucial piece of information as we move forward in applying the chain rule. Understanding how each variable changes with respect to others is essential for solving this type of problem. The simplicity of this derivative allows us to focus on the more complex derivatives that follow. It also highlights the importance of basic differentiation rules, which form the foundation for more advanced calculus techniques. By correctly identifying and calculating this derivative, we have taken the first step towards finding dy/dt.

2. du/ds: Given u = 1/s² = s⁻², we differentiate u with respect to s: du/ds = -2s⁻³ = -2/s³

   Here, we use the power rule of differentiation, which states that the derivative of xⁿ with respect to x is nx^(n-1). Applying this rule to s⁻², we get -2s^(-2-1) = -2s⁻³*. We then rewrite s⁻³ as 1/s³, resulting in du/ds = -2/s³. This derivative tells us how u changes with respect to s. For instance, if s increases, u will decrease, and vice versa. The negative sign indicates an inverse relationship between u and s. This derivative is crucial for linking the change in s to the change in u, which ultimately affects the change in y. It's important to pay attention to the sign and the power of s in the derivative, as they provide valuable information about the relationship between the variables. By correctly calculating du/ds, we have successfully linked u and s in our chain rule calculation.

3. ds/dt: Given s = 1 - t, we differentiate s with respect to t: ds/dt = -1

   This is another straightforward derivative. The derivative of a constant (1) is zero, and the derivative of -t with respect to t is -1. This means that for every unit increase in t, s decreases by 1 unit. This derivative is particularly important because it directly relates s to the independent variable t. The negative sign indicates that s and t have an inverse relationship; as t increases, s decreases. This simple derivative forms a critical link in the chain rule, allowing us to connect the change in t to the changes in s, u, and ultimately, y. Understanding this relationship is essential for finding dy/dt. The constant value of the derivative (-1) simplifies the subsequent calculations and provides a clear understanding of how s responds to changes in t.

Now, we substitute these derivatives back into the chain rule equation:

dy/dt = (3) * (-2/s³) * (-1)

dy/dt = 6/s³

Finally, we substitute s = 1 - t into the equation to express dy/dt in terms of t:

dy/dt = 6/(1 - t)³

This is the final derivative of y with respect to t. The expression 6/(1-t)³ shows how the rate of change of y with respect to t depends on the value of t. As t approaches 1, the denominator approaches zero, and the derivative approaches infinity, indicating a very rapid change in y. This result highlights the importance of considering the domain of the function and potential singularities. The process of substituting s = 1 - t is crucial for expressing the derivative in terms of the original independent variable, t. This makes the result more meaningful and directly relates the change in y to the change in t. By successfully performing this substitution, we have completed the solution and obtained the desired derivative. The final expression provides valuable insights into the relationship between y and t and demonstrates the power of the chain rule in solving complex differentiation problems.

In this article, we successfully found dy/dt given the relationships y = 3u + 1, u = 1/s², and s = 1 - t. We achieved this by systematically applying the chain rule, a fundamental concept in calculus. The chain rule allowed us to break down the complex dependency of y on t into smaller, manageable derivatives. We first calculated dy/du, du/ds, and ds/dt individually, and then combined them using the chain rule formula: dy/dt = (dy/du) * (du/ds) * (ds/dt). Finally, we substituted s = 1 - t into the resulting expression to obtain dy/dt in terms of t, which is dy/dt = 6/(1 - t)³. The solution demonstrates the power and versatility of the chain rule in handling composite functions. Understanding the chain rule is crucial for solving a wide range of problems in calculus and related fields. This exercise also highlights the importance of understanding functional dependencies and how variables interact within a system of equations. The step-by-step approach used in this article provides a clear and concise method for tackling similar problems. By breaking down the problem into smaller steps, we were able to manage the complexity and arrive at the correct solution. This problem serves as a valuable example of how to apply the chain rule effectively and reinforces the fundamental concepts of differentiation. The final result, dy/dt = 6/(1 - t)³, provides a clear understanding of how the rate of change of y with respect to t varies depending on the value of t. This solution not only provides the answer but also enhances our understanding of the underlying calculus principles.