How To Find The Derivative Of S = 4t^(5/4) + 9e^3 A Step-by-Step Guide
Introduction
In calculus, the derivative of a function measures the instantaneous rate of change of the function with respect to one of its variables. It is a fundamental concept with wide applications in various fields such as physics, engineering, economics, and computer science. Finding the derivative of a function is a core skill in calculus, and this article will guide you through the process of finding the derivative of the function s = 4t^(5/4) + 9e^3. We will break down the function into its components, apply the relevant differentiation rules, and arrive at the final derivative. Understanding how to find derivatives is crucial for anyone studying or working in quantitative fields, as it allows us to analyze how quantities change and interact. This article provides a step-by-step approach to ensure clarity and comprehension, making it accessible for both beginners and those looking to refresh their knowledge.
Understanding the Function
Before we dive into the process of finding the derivative, it is essential to understand the function we are working with. The function given is s = 4t^(5/4) + 9e^3. This function has two main terms: 4t^(5/4) and 9e^3. Let's break down each term individually to understand their structure and components.
The first term, 4t^(5/4), is a power function. A power function is a function in the form of ax^n, where 'a' is a constant coefficient, 'x' is the variable, and 'n' is a real number exponent. In our case, the constant coefficient is 4, the variable is 't', and the exponent is 5/4. Understanding the structure of this term is crucial because we will use the power rule of differentiation to find its derivative. The power rule is a fundamental rule in calculus that simplifies the process of differentiating power functions. It states that the derivative of x^n with respect to 'x' is nx^(n-1). We will apply this rule to 4t^(5/4), considering the coefficient and the exponent to find its derivative.
The second term, 9e^3, might appear complex at first glance, but it is a constant term. The number 'e' is the base of the natural logarithm, approximately equal to 2.71828. In the term 9e^3, 'e' is raised to the power of 3, and the result is multiplied by 9. Since 'e' is a constant, and 3 is a constant, e^3 is also a constant. Multiplying it by 9 still results in a constant value. Thus, 9e^3 is a constant term. The derivative of a constant term is always zero, a rule that simplifies the differentiation process. Recognizing this term as a constant is crucial because it immediately tells us that its derivative will not contribute to the overall derivative of the function.
By understanding the structure of each term in the function s = 4t^(5/4) + 9e^3, we set the stage for applying the appropriate differentiation rules. The power rule will be used for the first term, and the constant rule will be used for the second term. This breakdown is essential for a clear and accurate differentiation process.
Applying Differentiation Rules
Now that we have a clear understanding of the function s = 4t^(5/4) + 9e^3 and its components, we can proceed to apply the differentiation rules to find its derivative. The key rules we will use are the power rule and the constant rule. The power rule is essential for differentiating terms in the form of ax^n, while the constant rule is crucial for differentiating constant terms. We will apply these rules step by step to ensure clarity and accuracy.
Power Rule
The power rule states that if we have a term in the form of ax^n, where 'a' is a constant coefficient and 'n' is a real number exponent, the derivative of this term with respect to 'x' is given by nax^(n-1). This rule is widely used in calculus because it simplifies the differentiation of power functions. It involves multiplying the original exponent by the coefficient and then reducing the exponent by 1. This straightforward process makes it easy to differentiate terms like x^2, x^3, and even more complex terms like x^(5/4). Understanding and applying the power rule is a fundamental skill in calculus, allowing us to find derivatives of various polynomial and rational functions efficiently.
In our function, we have the term 4t^(5/4). Here, the coefficient 'a' is 4, and the exponent 'n' is 5/4. Applying the power rule, we multiply the exponent (5/4) by the coefficient (4), which gives us (5/4) * 4 = 5. Then, we reduce the exponent by 1, which means we calculate (5/4) - 1 = 5/4 - 4/4 = 1/4. Thus, the derivative of 4t^(5/4) with respect to 't' is 5t^(1/4). This step demonstrates the direct application of the power rule, transforming a power function into its derivative by adjusting the coefficient and exponent. This is a critical step in finding the overall derivative of the function.
Constant Rule
The constant rule is another fundamental rule in calculus that simplifies the differentiation process. It states that the derivative of a constant term is always zero. A constant term is a value that does not change with respect to the variable being differentiated. Examples of constant terms include numbers like 5, 10, or even expressions like π or e. Understanding this rule is crucial because it allows us to quickly identify and eliminate constant terms from the differentiation process. This significantly reduces the complexity of finding derivatives for functions that include constant terms.
In our function s = 4t^(5/4) + 9e^3, the second term 9e^3 is a constant term. As we discussed earlier, 'e' is the base of the natural logarithm, approximately equal to 2.71828, and e^3 is a constant value. Multiplying it by 9 still results in a constant value. Therefore, 9e^3 is a constant. According to the constant rule, the derivative of 9e^3 with respect to 't' is 0. This simplifies our calculation because we know that this term will not contribute to the derivative of the function. The constant rule is a powerful tool for simplifying differentiation, especially when dealing with more complex functions that include multiple constant terms.
Step-by-Step Differentiation
Having established the necessary differentiation rules, we can now proceed with the step-by-step differentiation of the function s = 4t^(5/4) + 9e^3. This process involves applying the power rule to the term 4t^(5/4) and the constant rule to the term 9e^3. By breaking down the differentiation into clear steps, we can ensure accuracy and understanding.
Step 1: Differentiate the first term
The first term of the function is 4t^(5/4). To differentiate this term, we apply the power rule. As we discussed earlier, the power rule states that the derivative of ax^n with respect to 'x' is nax^(n-1). In this case, 'a' is 4, and 'n' is 5/4. Applying the power rule, we multiply the exponent (5/4) by the coefficient (4), which gives us (5/4) * 4 = 5. Next, we reduce the exponent by 1, which means we calculate (5/4) - 1 = 5/4 - 4/4 = 1/4. Therefore, the derivative of 4t^(5/4) with respect to 't' is 5t^(1/4). This step involves a straightforward application of the power rule, transforming the power function into its derivative by adjusting the coefficient and exponent. The result, 5t^(1/4), is the derivative of the first term and will be a part of the overall derivative of the function.
Step 2: Differentiate the second term
The second term of the function is 9e^3. This term is a constant, as 'e' is the base of the natural logarithm (approximately 2.71828), and 9e^3 is a constant value. To differentiate this term, we apply the constant rule. The constant rule states that the derivative of a constant term is always zero. Therefore, the derivative of 9e^3 with respect to 't' is 0. This step is straightforward, as the constant rule simplifies the differentiation process by immediately identifying constant terms and their derivatives. The fact that the derivative of 9e^3 is 0 means that this term does not contribute to the overall derivative of the function.
Step 3: Combine the derivatives
Now that we have found the derivatives of each term individually, we need to combine them to find the overall derivative of the function s = 4t^(5/4) + 9e^3. The derivative of the first term, 4t^(5/4), is 5t^(1/4), and the derivative of the second term, 9e^3, is 0. To find the overall derivative, we simply add the derivatives of each term together. Therefore, the derivative of s with respect to 't', denoted as ds/dt, is 5t^(1/4) + 0. This simplifies to 5t^(1/4). This final step combines the results of the previous steps to provide the complete derivative of the function. The derivative 5t^(1/4) represents the instantaneous rate of change of the function 's' with respect to 't'.
Final Derivative
After applying the differentiation rules and following a step-by-step process, we have arrived at the final derivative of the function s = 4t^(5/4) + 9e^3. The derivative, denoted as ds/dt, is 5t^(1/4). This result is a crucial outcome, representing the instantaneous rate of change of the function 's' with respect to the variable 't'. Understanding and interpreting this derivative is essential for various applications in mathematics, physics, engineering, and other fields.
The derivative ds/dt = 5t^(1/4) tells us how the value of 's' changes as 't' changes. It is a function itself, and its value at a specific point 't' gives us the rate of change of 's' at that point. For example, if we want to know how 's' is changing when 't' is 1, we can substitute t = 1 into the derivative, giving us 5(1)^(1/4) = 5. This means that at t = 1, 's' is changing at a rate of 5 units per unit change in 't'. The derivative provides valuable information about the behavior of the original function, including its increasing and decreasing intervals, its maximum and minimum points, and its concavity.
In summary, the final derivative of the function s = 4t^(5/4) + 9e^3 is ds/dt = 5t^(1/4). This result was obtained by applying the power rule to the term 4t^(5/4) and the constant rule to the term 9e^3. Understanding the process and the result is fundamental for calculus and its applications.
Applications of Derivatives
The derivative, as we found for the function s = 4t^(5/4) + 9e^3, is a fundamental concept in calculus with a wide array of applications across various fields. Understanding the applications of derivatives helps to appreciate their significance and practical relevance. Derivatives are used in physics to calculate velocity and acceleration, in engineering to optimize designs, in economics to analyze marginal cost and revenue, and in computer science for machine learning algorithms, among many other applications. Let's explore some key applications of derivatives to illustrate their importance.
Physics
In physics, derivatives are essential for describing motion. If we have a function that gives the position of an object as a function of time, the first derivative of this function gives the velocity of the object, and the second derivative gives the acceleration. For example, if s(t) represents the position of an object at time 't', then the velocity v(t) is given by the first derivative ds/dt, and the acceleration a(t) is given by the second derivative d2s/dt2 or dv/dt. These concepts are fundamental in classical mechanics and are used to analyze the motion of projectiles, the oscillations of pendulums, and the dynamics of systems in general. The ability to calculate velocity and acceleration from position functions is crucial for understanding and predicting the behavior of physical systems.
Engineering
In engineering, derivatives are used extensively for optimization. Engineers often need to design structures, circuits, or processes that maximize efficiency or minimize cost. Derivatives provide a powerful tool for finding the maximum or minimum values of functions, which is essential for optimization problems. For instance, in structural engineering, derivatives can be used to determine the shape of a bridge that minimizes stress and strain. In electrical engineering, derivatives can be used to design circuits that maximize power output or minimize energy consumption. In chemical engineering, derivatives can be used to optimize reaction rates and yields. The use of derivatives in optimization allows engineers to create more efficient, reliable, and cost-effective designs.
Economics
In economics, derivatives are used to analyze marginal concepts. Marginal cost, marginal revenue, and marginal profit are all key concepts in economic analysis, and they are defined using derivatives. Marginal cost, for example, is the derivative of the total cost function and represents the additional cost of producing one more unit of a product. Similarly, marginal revenue is the derivative of the total revenue function and represents the additional revenue from selling one more unit. Marginal profit is the derivative of the total profit function. By analyzing these marginal concepts, economists can make informed decisions about production levels, pricing strategies, and resource allocation. Derivatives provide a precise and quantitative way to understand the impact of small changes in economic variables.
Machine Learning
In computer science, particularly in machine learning, derivatives play a crucial role in training algorithms. Many machine learning algorithms, such as neural networks, rely on optimization techniques to learn from data. These algorithms adjust their parameters to minimize a cost function, which measures the difference between the algorithm's predictions and the actual values. Derivatives are used to calculate the gradient of the cost function, which indicates the direction of steepest descent. By following the gradient, the algorithm can iteratively adjust its parameters to find the minimum of the cost function, thereby improving its performance. The process of gradient descent is a fundamental technique in machine learning, and it relies heavily on the concept of derivatives. Without derivatives, many machine learning algorithms would not be able to learn effectively.
Conclusion
In conclusion, finding the derivative of the function s = 4t^(5/4) + 9e^3 involves applying fundamental rules of calculus, specifically the power rule and the constant rule. We systematically broke down the function, differentiated each term, and combined the results to arrive at the final derivative, ds/dt = 5t^(1/4). This process underscores the importance of understanding the underlying principles of differentiation and how they apply to various functions.
The derivative, as we've seen, is not just an abstract mathematical concept but a powerful tool with widespread applications. From physics and engineering to economics and computer science, derivatives enable us to analyze rates of change, optimize processes, and make informed decisions. The applications discussed highlight the versatility and practical relevance of derivatives in solving real-world problems.
Mastering the techniques of differentiation is crucial for anyone pursuing studies or careers in quantitative fields. The ability to find derivatives allows for a deeper understanding of how systems and processes behave and how they can be optimized. Whether you are calculating velocities and accelerations, designing efficient structures, analyzing economic trends, or training machine learning models, the derivative is an indispensable tool.
By understanding and applying differentiation rules, we can unlock insights into the dynamic relationships between variables and gain a more comprehensive understanding of the world around us. The function s = 4t^(5/4) + 9e^3 serves as a valuable example for illustrating the process of finding derivatives, and the broader applications of derivatives demonstrate their fundamental role in numerous disciplines. The journey of mastering calculus begins with understanding the basics, and derivatives are a cornerstone of that foundation. This article has provided a comprehensive guide to finding the derivative of a specific function and a glimpse into the vast applications of this essential concept.