Derivative Of F(x) = 3 Sin-1(x^4) And Evaluation At X = 0.6
Introduction
In the realm of calculus, derivatives play a pivotal role in understanding the rate of change of functions. This article delves into the process of finding the derivative of a specific function, $f(x) = 3 \sin{-1}(x4)$, and subsequently evaluating this derivative at a particular point, $x = 0.6$. We will navigate through the application of fundamental differentiation rules, particularly the chain rule and the derivative of the inverse sine function, to arrive at the solution. Furthermore, we will explore the significance of such calculations in various scientific and engineering applications.
Determining the Derivative of $f(x) = 3 \sin{-1}(x4)$
To embark on this mathematical journey, we first need to compute the derivative of the given function, $f(x) = 3 \sin{-1}(x4)$. The function involves a composition of functions, namely the inverse sine function and a power function. This necessitates the application of the chain rule, a cornerstone of differential calculus. The chain rule states that the derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function.
In our case, the outer function is $3\sin^-1}(u)$, where $u = x^4$ is the inner function. Let's first recall the derivative of the inverse sine function{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}}$. Applying this, the derivative of the outer function with respect to $u$ is $3 \cdot \frac{1}{\sqrt{1 - u^2}}$. Next, we need to find the derivative of the inner function, $u = x^4$, with respect to $x$. Using the power rule, we find that $\frac{du}{dx} = 4x^3$.
Now, we can apply the chain rule: $\frac{df}{dx} = \frac{d}{du}(3\sin^{-1}(u)) \cdot \frac{du}{dx} = 3 \cdot \frac{1}{\sqrt{1 - u^2}} \cdot 4x^3$. Substituting $u = x^4$ back into the expression, we get $f'(x) = \frac{12x^3}{\sqrt{1 - (x4)2}} = \frac{12x^3}{\sqrt{1 - x^8}}$. This expression represents the derivative of the function $f(x)$ with respect to $x$. The derivative gives us the instantaneous rate of change of the function at any given point $x$. Understanding this rate of change is crucial in various applications, such as optimization problems, physics simulations, and economic modeling.
Evaluating $f'(0.6)$
Having found the derivative $f'(x) = \frac{12x^3}{\sqrt{1 - x^8}}$, our next task is to evaluate it at $x = 0.6$. This involves substituting $0.6$ for $x$ in the expression and performing the necessary calculations. This process provides us with the specific rate of change of the function at the point $x = 0.6$, offering valuable insight into the function's behavior at that particular location.
Substituting $x = 0.6$ into the derivative, we get $f'(0.6) = \frac12(0.6)^3}{\sqrt{1 - (0.6)^8}}$. Now, we need to compute the numerical value. First, we calculate $(0.6)^3 = 0.216$ and $(0.6)^8 \approx 0.01679616$. Then, we have $12(0.6)^3 = 12 \cdot 0.216 = 2.592$. Next, we compute the denominator \approx \sqrt1 - 0.01679616} \approx \sqrt{0.98320384} \approx 0.991566$. Finally, we divide the numerator by the denominator{0.991566} \approx 2.614$. Therefore, the value of the derivative at $x = 0.6$ is approximately $2.614$, rounded to three decimal places.
This numerical result signifies the slope of the tangent line to the graph of $f(x)$ at the point where $x = 0.6$. It indicates the instantaneous rate at which the function's value is changing at that specific input. In practical applications, this could represent, for example, the velocity of an object at a certain time or the marginal cost of production at a particular output level. Understanding the derivative's value at specific points is fundamental to making informed decisions and predictions in various fields.
Significance and Applications of Derivatives
Derivatives are not merely abstract mathematical concepts; they are powerful tools with far-reaching applications across diverse fields. Their ability to quantify rates of change makes them indispensable in physics, engineering, economics, and computer science, among others. From optimizing designs to modeling complex systems, derivatives provide essential insights into the behavior of functions and the phenomena they represent.
In physics, derivatives are used to describe velocity and acceleration. The velocity of an object is the derivative of its position with respect to time, while acceleration is the derivative of velocity with respect to time. These concepts are fundamental to understanding motion and are used in everything from classical mechanics to astrophysics. For instance, understanding the trajectory of a projectile or the orbit of a planet relies heavily on the principles of differentiation.
In engineering, derivatives are crucial for optimization problems. Engineers often need to design structures or systems that maximize efficiency or minimize cost. This often involves finding the maximum or minimum of a function, which can be done using derivatives. For example, designing a bridge that can withstand the most weight or creating an electrical circuit that uses the least amount of power involves the application of differential calculus.
In economics, derivatives are used to analyze marginal cost and marginal revenue. Marginal cost is the derivative of the total cost function, and marginal revenue is the derivative of the total revenue function. These concepts are used to determine the optimal level of production for a company. Understanding how costs and revenues change with production levels allows businesses to make strategic decisions about pricing and output.
In computer science, derivatives are used in machine learning algorithms. Many machine learning algorithms involve minimizing a cost function, which can be done using gradient descent, a method that relies on derivatives. For instance, training a neural network to recognize images or predict stock prices involves iteratively adjusting parameters based on the gradient of a cost function.
The calculation we performed in this article, finding and evaluating the derivative of $f(x) = 3 \sin{-1}(x4)$, is a microcosm of the broader applications of calculus. The ability to differentiate complex functions and interpret the results is a valuable skill for anyone working in a quantitative field.
Conclusion
In this article, we successfully computed the derivative of the function $f(x) = 3 \sin{-1}(x4)$ using the chain rule and the derivative of the inverse sine function. We found that $f'(x) = \frac{12x^3}{\sqrt{1 - x^8}}$. Furthermore, we evaluated the derivative at $x = 0.6$, obtaining an approximate value of $2.614$. This process highlights the power of calculus in analyzing the behavior of functions and quantifying their rates of change. The significance of derivatives extends far beyond theoretical mathematics, impacting numerous disciplines and real-world applications. From physics and engineering to economics and computer science, derivatives provide essential tools for modeling, optimizing, and understanding complex systems. The ability to find and interpret derivatives is a fundamental skill for anyone working in a quantitative field, enabling informed decision-making and problem-solving in a wide range of contexts. Understanding the concepts and techniques discussed in this article provides a solid foundation for further exploration of calculus and its applications.
