Derivative Of X/2 Sqrt(a^2-x^2) + A^2/2 Sin^-1(x/a) Equals Sqrt(a^2-x^2)

In the realm of calculus, derivatives play a pivotal role in understanding the rate of change of functions. This article delves into the intricate process of finding the derivative of a specific expression involving trigonometric functions and square roots. Our primary focus is to demonstrate why the derivative of the expression $\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$ equals $\sqrt{a^2-x^2}$. This exploration will not only reinforce your understanding of differentiation rules but also shed light on the interplay between algebraic and trigonometric concepts.

Differentiation Fundamentals: Setting the Stage

Before we embark on the main derivation, it's crucial to lay a solid foundation by revisiting some fundamental differentiation rules. These rules will serve as our guiding principles throughout the process. At the heart of differential calculus lies the concept of a derivative, which quantifies the instantaneous rate of change of a function. Mastering the rules of differentiation is akin to possessing the keys to unlock the secrets of how functions behave. These rules enable us to systematically determine the rate at which a function's output changes with respect to its input, a cornerstone of calculus and its applications.

The Product Rule: Navigating Multiplication

When faced with the derivative of a product of two functions, the product rule comes to our rescue. This rule elegantly states that the derivative of the product of two functions is the sum of the first function times the derivative of the second function and the second function times the derivative of the first function. In mathematical notation, if we have two functions, u(x) and v(x), then the derivative of their product, d/dx [u(x)v(x)], is given by u'(x)v(x) + u(x)v'(x). The product rule is indispensable when dealing with expressions where two functions are multiplied together, allowing us to dissect the derivative into manageable components. This rule is a cornerstone of calculus, allowing us to tackle complex expressions involving the multiplication of functions with confidence and precision.

The Chain Rule: Unraveling Composite Functions

Composite functions, where one function is nested inside another, require a special approach. This approach is precisely what the chain rule provides. The chain rule dictates 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. Symbolically, if we have a composite function f(g(x)), then its derivative, d/dx [f(g(x))], is expressed as f'(g(x)) * g'(x). This rule is fundamental for handling compositions of functions, allowing us to systematically break down the differentiation process into manageable steps. By applying the chain rule, we can unravel the layers of a composite function, determining its derivative with accuracy and efficiency.

Derivatives of Key Functions: Our Building Blocks

To effectively tackle the derivative in question, we need to have a firm grasp of the derivatives of some key functions. The derivative of sin⁻¹(x), also known as the inverse sine function, is given by 1/√(1-x²). This derivative is a fundamental result in calculus and is frequently encountered in various applications. The derivative of √x, the square root function, is 1/(2√x). This result is essential for handling expressions involving square roots and is a cornerstone of differentiation techniques. Armed with these derivatives, we are well-equipped to tackle the derivative of the given expression. These derivatives serve as essential building blocks, enabling us to dissect the expression and apply the appropriate rules to find its derivative with precision.

Dissecting the Expression: A Strategic Approach

Our mission is to find the derivative of the expression $\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$. To do this effectively, we'll break down the expression into smaller, more manageable parts. This strategic approach will allow us to apply the differentiation rules systematically and minimize the chances of errors. By dissecting the expression, we can focus on each component individually, making the overall differentiation process more streamlined and efficient. This approach not only simplifies the task at hand but also enhances our understanding of the expression's structure and behavior.

Term 1: Applying the Product Rule

The first term, $\frac{x}{2} \sqrt{a^2-x^2}$, is a product of two functions: $\frac{x}{2}$ and $\sqrt{a^2-x^2}$. This immediately calls for the application of the product rule. Let's define u(x) = x/2 and v(x) = √(a² - x²). Then, the derivative of u(x), u'(x), is simply 1/2. To find the derivative of v(x), v'(x), we'll need to employ the chain rule. The product rule is a fundamental tool in calculus, allowing us to differentiate expressions that involve the multiplication of functions. By recognizing the product structure of the first term, we can strategically apply the product rule to break down the differentiation process into manageable steps. This systematic approach ensures that we account for the interplay between the two functions and accurately determine the derivative of their product.

Chain Rule in Action: Differentiating the Square Root

The function v(x) = √(a² - x²) is a composite function. We can think of it as the square root function applied to the expression (a² - x²). Using the chain rule, we first differentiate the outer function, the square root, with respect to its argument, and then multiply by the derivative of the inner function, (a² - x²). The derivative of the square root function is 1/(2√x), and the derivative of (a² - x²) is -2x. Combining these, we get v'(x) = (1/(2√(a² - x²))) * (-2x) = -x/√(a² - x²). The chain rule is essential for differentiating composite functions, allowing us to systematically account for the nested structure of the function. By carefully applying the chain rule, we can accurately determine the derivative of the square root term, ensuring that we capture the impact of both the outer square root function and the inner expression (a² - x²).

Putting it Together: The Derivative of Term 1

Now, we can apply the product rule: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Substituting the derivatives we found, we get: d/dx [$\frac{x}{2} \sqrt{a^2-x^2}$] = (1/2)√(a² - x²) + (x/2)(-x/√(a² - x²)). Simplifying this expression, we get: (√(a² - x²)/2) - (x²/(2√(a² - x²))). This result represents the derivative of the first term in the expression. The product rule, combined with the chain rule, allows us to systematically differentiate the first term, accounting for the interplay between the linear term and the square root term. By carefully applying these rules and simplifying the resulting expression, we arrive at the derivative of the first term, which will contribute to the overall derivative of the expression.

Term 2: Differentiating the Inverse Sine Function

The second term, $\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$, involves the inverse sine function. The derivative of sin⁻¹(x) is 1/√(1-x²). However, we have sin⁻¹(x/a), so we'll need to use the chain rule again. Let's consider the constant factor a²/2 separately for now. The inverse sine function, also known as arcsine, is a fundamental trigonometric function whose derivative is a crucial result in calculus. By understanding the derivative of the inverse sine function, we can effectively tackle the differentiation of the second term, which involves a composition of the inverse sine function with a linear term. This step is essential for determining the overall derivative of the expression.

Applying the Chain Rule: Inverse Sine with a Twist

Let u = x/a. Then, we have sin⁻¹(u). The derivative of sin⁻¹(u) with respect to u is 1/√(1-u²). The derivative of u = x/a with respect to x is 1/a. Applying the chain rule, the derivative of sin⁻¹(x/a) with respect to x is (1/√(1-(x/a)²)) * (1/a). Simplifying, we get 1/√(a² - x²). The chain rule is once again instrumental in handling the composite function involving the inverse sine function and the linear term x/a. By carefully applying the chain rule, we can accurately determine the derivative of this composite function, ensuring that we account for the impact of both the inverse sine function and the linear term. This step is crucial for obtaining the complete derivative of the second term.

Accounting for the Constant: The Derivative of Term 2

Now, we multiply by the constant factor a²/2: d/dx [$\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$] = (a²/2) * (1/√(a² - x²)) = a²/(2√(a² - x²)). This is the derivative of the second term. The constant factor scales the derivative of the inverse sine function, contributing to the overall derivative of the expression. By accounting for the constant factor, we ensure that the derivative of the second term is accurately determined, reflecting the influence of the constant on the rate of change of the term.

The Grand Finale: Combining the Derivatives

Now that we have the derivatives of both terms, we can add them together to find the derivative of the entire expression: d/dx [$\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$] = (√(a² - x²)/2) - (x²/(2√(a² - x²))) + a²/(2√(a² - x²)). To simplify this, we can combine the fractions: (√(a² - x²)/2) + (a² - x²)/(2√(a² - x²)). The combination of the derivatives is the final step in determining the derivative of the entire expression. By adding the derivatives of the individual terms, we capture the overall rate of change of the expression. This step requires careful attention to algebraic manipulation and simplification to arrive at the most concise form of the derivative.

Simplifying the Expression: Unveiling the Result

Further simplification yields: ((a² - x²)/(2√(a² - x²))) + ((a² - x²)/(2√(a² - x²))) = (a² - x²)/(√(a² - x²)). Finally, we can simplify this to √(a² - x²). Thus, we have shown that the derivative of the given expression is indeed √(a² - x²). The simplification process is crucial for arriving at the final, most concise form of the derivative. By carefully manipulating the algebraic expression, we can eliminate unnecessary terms and reveal the underlying structure of the derivative. In this case, the simplification leads to the elegant result √(a² - x²), demonstrating the power of calculus to unveil hidden relationships within mathematical expressions.

Conclusion: A Triumph of Calculus

In this article, we have successfully demonstrated that the derivative of $\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$ is $\sqrt{a^2-x^2}$. This journey involved a careful application of the product rule, chain rule, and the derivatives of key functions. This example highlights the power and elegance of calculus in unraveling the intricacies of mathematical expressions. The conclusion summarizes the journey we have undertaken, highlighting the key steps and the final result. This serves to reinforce the understanding of the process and the significance of the result. By successfully demonstrating the derivative of the given expression, we have showcased the power of calculus to tackle complex mathematical problems and reveal underlying relationships.

This exploration not only solidifies our understanding of differentiation techniques but also underscores the interconnectedness of various mathematical concepts. The interplay between algebraic manipulation, trigonometric functions, and calculus principles is a testament to the beauty and coherence of mathematics. This example serves as a valuable exercise in applying fundamental calculus rules and provides a deeper appreciation for the elegance and power of mathematical reasoning. By working through this derivation, we have honed our skills in differentiation and gained a greater understanding of the relationships between different mathematical concepts. This knowledge will serve as a valuable foundation for tackling more complex problems in calculus and related fields.