Given F(x)=ax+1/(6x-5). If F'(1/6)=-21/16, What Is The Value Of A?

In the realm of calculus, understanding derivatives is crucial for analyzing the behavior of functions. Derivatives provide insights into the rate of change of a function, allowing us to determine its increasing or decreasing nature, find critical points, and solve optimization problems. This article delves into the concept of derivatives and applies it to a specific problem: finding the value of 'a' in the function f(x) = ax + 1/6x - 5, given that f'(1/6) = -21/16. This problem serves as an excellent example of how derivatives can be used to solve for unknown parameters in a function.

The Essence of Derivatives

At its core, the derivative of a function measures its instantaneous rate of change at a particular point. Imagine driving a car; your speedometer tells you your speed at any given moment. This is analogous to the derivative – it tells us how the function's output is changing with respect to its input at a specific point. Mathematically, the derivative of a function f(x) is denoted as f'(x) and is defined as the limit of the difference quotient as the change in x approaches zero:

f'(x) = lim (h->0) [f(x + h) - f(x)] / h

This formula might seem daunting at first, but it simply represents the slope of the line tangent to the function's graph at a specific point. The tangent line is the best linear approximation of the function at that point, and its slope indicates how steeply the function is increasing or decreasing. The derivative, therefore, provides a powerful tool for understanding the local behavior of a function.

Differentiation Rules: Tools for Finding Derivatives

Calculating derivatives using the limit definition can be cumbersome, especially for complex functions. Fortunately, a set of rules simplifies the process of differentiation. These rules allow us to find the derivatives of various types of functions without resorting to the limit definition every time. Some of the fundamental differentiation rules include:

• Power Rule: If f(x) = x^n, then f'(x) = nx^(n-1)
• Constant Multiple Rule: If f(x) = cf(x), where c is a constant, then f'(x) = cf'(x)
• Sum/Difference Rule: If f(x) = u(x) ± v(x), then f'(x) = u'(x) ± v'(x)
• Product Rule: If f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x)
• Quotient Rule: If f(x) = u(x) / v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2
• Chain Rule: If f(x) = g(h(x)), then f'(x) = g'(h(x))h'(x)

These rules, when applied correctly, can significantly simplify the process of finding derivatives. In the given problem, we will primarily use the power rule, constant multiple rule, sum/difference rule, and quotient rule.

Applying Derivatives to Solve for 'a'

Now, let's tackle the problem at hand. We are given the function f(x) = ax + 1/(6x - 5) and the information that f'(1/6) = -21/16. Our goal is to find the value of the constant 'a'. To do this, we will first find the derivative of f(x) and then use the given condition to solve for 'a'.

The function f(x) can be viewed as the sum of two terms: ax and 1/(6x - 5). To find the derivative, we will differentiate each term separately and then add the results. The derivative of ax with respect to x is simply 'a' (using the power rule and constant multiple rule). To find the derivative of 1/(6x - 5), we will use the quotient rule. Let u(x) = 1 and v(x) = 6x - 5. Then, u'(x) = 0 and v'(x) = 6. Applying the quotient rule, we get:

d/dx [1/(6x - 5)] = [0(6x - 5) - 1(6)] / (6x - 5)^2 = -6 / (6x - 5)^2

Therefore, the derivative of f(x) is:

f'(x) = a - 6 / (6x - 5)^2

Now, we are given that f'(1/6) = -21/16. We can substitute x = 1/6 into the expression for f'(x) and set it equal to -21/16:

f'(1/6) = a - 6 / (6(1/6) - 5)^2 = -21/16

Simplifying the expression, we get:

a - 6 / (1 - 5)^2 = -21/16

a - 6 / 16 = -21/16

Now, we can solve for 'a':

a = -21/16 + 6/16

a = -15/16

Therefore, the value of 'a' is -15/16. This solution demonstrates how the concept of derivatives, combined with differentiation rules, can be applied to solve for unknown parameters in a function.

Conclusion: The Power of Derivatives

This problem highlights the power and versatility of derivatives in calculus. By understanding the concept of derivatives and mastering differentiation rules, we can analyze the behavior of functions, solve optimization problems, and, as demonstrated in this example, find unknown parameters. The derivative is a fundamental tool in mathematics and has applications in various fields, including physics, engineering, economics, and computer science. Mastering derivatives is an essential step for anyone seeking a deeper understanding of these fields. In summary, the solution to the problem f(x) = ax + 1/(6x - 5) with f'(1/6) = -21/16 involves finding the derivative using the quotient rule, substituting the given value, and solving for 'a', which yields a = -15/16. The application of calculus principles allows us to unravel the mysteries of functions and their rates of change.

Further Exploration of Derivatives

To further solidify your understanding of derivatives, consider exploring the following topics:

• Applications of Derivatives: Investigate how derivatives are used to find maximum and minimum values of functions (optimization), analyze the concavity of curves, and determine inflection points.
• Higher-Order Derivatives: Learn about second derivatives, third derivatives, and so on, and how they relate to the rate of change of the rate of change.
• Implicit Differentiation: Explore how to find derivatives of implicitly defined functions.
• Related Rates Problems: Practice solving problems that involve finding the rates of change of related quantities.

By delving deeper into these topics, you will gain a more comprehensive understanding of the power and applications of derivatives in calculus and beyond. The journey of understanding calculus is a rewarding one, and derivatives are a crucial stepping stone on that path. Remember, practice makes perfect, so work through numerous examples and problems to hone your skills. The more you practice, the more intuitive the concepts will become, and the more confident you will be in your ability to apply them.