Finding The Derivative Of F(x) = √((x² - 2ax) / (a² - 2ab)) A Calculus Guide

Jul 10, 2025 by ADMIN 77 views

Finding the Derivative of a Radical Function: A Step-by-Step Guide

In this article, we will delve into the process of finding the derivative of a complex function involving radicals and algebraic expressions. Specifically, we will focus on the function f(x) = √((x² - 2ax) / (a² - 2ab)). This problem often appears in calculus courses and requires a solid understanding of differentiation rules, including the quotient rule, chain rule, and power rule. By breaking down the problem into manageable steps, we will provide a clear and concise guide to finding f'(x).

Understanding the Function

Before we dive into the differentiation process, it’s crucial to understand the function we’re dealing with. The function f(x) = √((x² - 2ax) / (a² - 2ab)) involves a square root of a quotient. This means we will need to apply several differentiation rules in a specific order to arrive at the correct derivative. Let's first rewrite the function to make it easier to differentiate. We can express the square root as a power of 1/2, giving us f(x) = ((x² - 2ax) / (a² - 2ab))^(1/2). This form is more amenable to applying the chain rule, which is essential when dealing with composite functions. Additionally, notice that the denominator (a² - 2ab) is a constant with respect to x, which will simplify our calculations later. Recognizing such constants is a key step in efficient differentiation. The numerator (x² - 2ax) is a polynomial, which we can easily differentiate using the power rule. Our strategy will involve applying the chain rule to the entire function, the quotient rule (if necessary), and the power rule to the individual terms. By carefully applying each rule, we will systematically find the derivative f'(x).

Applying Differentiation Rules

Now, let's embark on the journey of finding the derivative. Our function is f(x) = √((x² - 2ax) / (a² - 2ab)), which we've rewritten as f(x) = ((x² - 2ax) / (a² - 2ab))^(1/2). The first rule we'll employ is the chain rule. The chain rule states that if we have a composite function, the derivative of the outer function is multiplied by the derivative of the inner function. In this case, the outer function is u^(1/2), where u = (x² - 2ax) / (a² - 2ab), and the inner function is (x² - 2ax) / (a² - 2ab). Applying the chain rule, we get:

f'(x) = (1/2) * ((x² - 2ax) / (a² - 2ab))^(-1/2) * d/dx [(x² - 2ax) / (a² - 2ab)].

Next, we need to find the derivative of the inner function (x² - 2ax) / (a² - 2ab). Here, we can treat (a² - 2ab) as a constant denominator. Thus, we only need to differentiate the numerator (x² - 2ax) with respect to x. The derivative of x² is 2x, and the derivative of -2ax is -2a. Therefore, d/dx [x² - 2ax] = 2x - 2a. Since (a² - 2ab) is a constant, we can write:

d/dx [(x² - 2ax) / (a² - 2ab)] = (2x - 2a) / (a² - 2ab).

Now, we substitute this back into our expression for f'(x):

f'(x) = (1/2) * ((x² - 2ax) / (a² - 2ab))^(-1/2) * ((2x - 2a) / (a² - 2ab)).

This step combines the chain rule with the basic differentiation rules for polynomials. We've successfully differentiated the function piece by piece, making sure to keep track of each step.

Simplifying the Derivative

Our next task is to simplify the derivative we've obtained. The expression currently looks like this:

f'(x) = (1/2) * ((x² - 2ax) / (a² - 2ab))^(-1/2) * ((2x - 2a) / (a² - 2ab)).

First, let's rewrite the term with the negative exponent. A negative exponent means we take the reciprocal of the base and raise it to the positive exponent. So, ((x² - 2ax) / (a² - 2ab))^(-1/2) becomes ((a² - 2ab) / (x² - 2ax))^(1/2). Substituting this back into our expression, we get:

f'(x) = (1/2) * ((a² - 2ab) / (x² - 2ax))^(1/2) * ((2x - 2a) / (a² - 2ab)).

Now, we can combine the terms and simplify further. Notice that we have a factor of (1/2) and a factor of (2x - 2a) in the numerator. We can factor out a 2 from (2x - 2a), which gives us 2(x - a). Then, the (1/2) and the 2 will cancel each other out. Our expression now looks like:

f'(x) = ((a² - 2ab) / (x² - 2ax))^(1/2) * ((x - a) / (a² - 2ab)).

We can rewrite the term ((a² - 2ab) / (x² - 2ax))^(1/2) as √(a² - 2ab) / √(x² - 2ax). This allows us to express f'(x) as:

f'(x) = (√(a² - 2ab) / √(x² - 2ax)) * ((x - a) / (a² - 2ab)).

Now, we can combine the fractions:

f'(x) = ( (x - a) * √(a² - 2ab) ) / ( √(x² - 2ax) * (a² - 2ab) ).

To simplify further, we can rewrite (a² - 2ab) as √(a² - 2ab) * √(a² - 2ab). This will allow us to cancel out one of the √(a² - 2ab) terms in the numerator and denominator:

f'(x) = ( (x - a) * √(a² - 2ab) ) / ( √(x² - 2ax) * √(a² - 2ab) * √(a² - 2ab) ).

Cancelling out the common term, we get:

f'(x) = (x - a) / ( √(x² - 2ax) * √(a² - 2ab) ).

Finally, we can combine the square roots in the denominator:

f'(x) = (x - a) / ( √( (x² - 2ax) * (a² - 2ab) ) ).

This simplified form of the derivative is much cleaner and easier to work with. It showcases the power of algebraic manipulation in simplifying complex expressions. By carefully simplifying, we reduce the chances of making errors in subsequent calculations and gain a clearer understanding of the function's behavior.

Final Result

After meticulously applying the chain rule, quotient rule (implicitly), and power rule, and after several steps of simplification, we arrive at the final form of the derivative:

f'(x) = (x - a) / ( √( (x² - 2ax) * (a² - 2ab) ) ).

This result provides us with the instantaneous rate of change of the function f(x) = √((x² - 2ax) / (a² - 2ab)) at any point x in its domain. The derivative is a crucial tool in calculus, allowing us to analyze the function's behavior, find its critical points, and determine intervals of increase and decrease. The expression we've derived might seem complex, but it encapsulates all the necessary information about the rate of change of the original function.

To recap, we started with a function involving a square root and a quotient. We rewrote the function using exponent notation to make it easier to differentiate. We then applied the chain rule, followed by the differentiation of the inner function. We simplified the resulting expression by combining terms, factoring, and using properties of exponents and radicals. This step-by-step approach is essential when dealing with complex functions, as it breaks the problem down into manageable parts and reduces the likelihood of errors. The final derivative, f'(x) = (x - a) / ( √( (x² - 2ax) * (a² - 2ab) ) ), is the culmination of these efforts and provides valuable insights into the behavior of the original function.

In conclusion, finding the derivative of f(x) = √((x² - 2ax) / (a² - 2ab)) involved a combination of differentiation rules and algebraic simplification. The final result, f'(x) = (x - a) / ( √( (x² - 2ax) * (a² - 2ab) ) ), represents the instantaneous rate of change of the function. This exercise highlights the importance of mastering differentiation techniques and algebraic manipulation in calculus. Understanding these concepts allows us to tackle complex problems and gain a deeper understanding of mathematical functions.