How To Differentiate (secx - Tanx)/(secx + Tanx) With Respect To X? How To Differentiate Ln((1 - 3x²)/(1 + 3x²))^(1/2) With Respect To X?

by ADMIN 138 views

In the realm of calculus, differentiation stands as a fundamental operation, revealing the rate at which a function's output changes with respect to its input. This process is pivotal in various fields, including physics, engineering, economics, and computer science, enabling the modeling and analysis of dynamic systems and optimization problems. Among the myriad functions encountered in calculus, trigonometric and logarithmic functions hold significant importance. Trigonometric functions, such as sine, cosine, tangent, secant, cosecant, and cotangent, describe the relationships between angles and sides in triangles and are essential in modeling periodic phenomena. Logarithmic functions, the inverses of exponential functions, are crucial in solving equations involving exponents and analyzing growth and decay processes. This comprehensive guide delves into the intricacies of differentiating trigonometric and logarithmic functions, providing a step-by-step approach to solving complex problems. We will explore various techniques, including the quotient rule, chain rule, and properties of logarithms, to tackle differentiation challenges effectively. Grasping these concepts is not merely an academic exercise but a crucial skill for anyone seeking to apply mathematical principles to real-world scenarios. This guide aims to equip you with the knowledge and skills necessary to confidently differentiate trigonometric and logarithmic functions, fostering a deeper understanding of calculus and its applications. As we embark on this journey, remember that practice is the key to mastery. Work through the examples provided, and don't hesitate to explore additional problems to solidify your understanding. Calculus is a journey of discovery, and with each step, you'll unlock new insights into the mathematical world.

Differentiating (sec x - tan x) / (sec x + tan x)

In this section, we embark on a journey to differentiate the trigonometric function (sec x - tan x) / (sec x + tan x). This task necessitates a robust understanding of differentiation rules, particularly the quotient rule and the derivatives of trigonometric functions. The quotient rule, a cornerstone of calculus, provides a systematic method for differentiating functions expressed as the ratio of two differentiable functions. It states that the derivative of u(x) / v(x) is given by (v(x)u'(x) - u(x)v'(x)) / (v(x))^2, where u'(x) and v'(x) represent the derivatives of u(x) and v(x), respectively. To effectively apply the quotient rule, we must first identify the numerator and denominator of our function. In this case, u(x) = sec x - tan x and v(x) = sec x + tan x. Next, we need to determine the derivatives of sec x and tan x. Recall that the derivative of sec x is sec x tan x, and the derivative of tan x is sec^2 x. Now, we can apply the quotient rule to find the derivative of (sec x - tan x) / (sec x + tan x). Let's break down the process step-by-step. First, we find the derivatives of the numerator and denominator: u'(x) = sec x tan x - sec^2 x and v'(x) = sec x tan x + sec^2 x. Next, we apply the quotient rule formula: d/dx [(sec x - tan x) / (sec x + tan x)] = [(sec x + tan x)(sec x tan x - sec^2 x) - (sec x - tan x)(sec x tan x + sec^2 x)] / (sec x + tan x)^2. The next step is to simplify the expression. This involves expanding the products and combining like terms. After careful algebraic manipulation, we arrive at the simplified form: -2 sec^2 x tan x / (sec x + tan x)^2. This is the derivative of (sec x - tan x) / (sec x + tan x) with respect to x. The journey of differentiation doesn't end here. We can further simplify the expression by utilizing trigonometric identities. By expressing sec x as 1/cos x and tan x as sin x/cos x, we can rewrite the derivative in terms of sine and cosine functions. This often leads to a more compact and insightful representation of the derivative. Throughout this process, meticulous attention to detail is paramount. Each step, from identifying the quotient rule to simplifying the expression, requires careful execution to avoid errors. The final result, -2 sec^2 x tan x / (sec x + tan x)^2, provides valuable information about the rate of change of the original function. It tells us how the function (sec x - tan x) / (sec x + tan x) varies as x changes. This knowledge is crucial in various applications, such as analyzing the behavior of trigonometric functions in different contexts.

Step-by-step Differentiation of (sec x - tan x) / (sec x + tan x)

  1. Identify the Quotient Rule: Recognize that the function is a ratio of two expressions, making the quotient rule the appropriate differentiation technique.
  2. Define u(x) and v(x): Let u(x) = sec x - tan x and v(x) = sec x + tan x.
  3. Find u'(x) and v'(x): Calculate the derivatives of u(x) and v(x). Recall that d/dx (sec x) = sec x tan x and d/dx (tan x) = sec^2 x. Therefore, u'(x) = sec x tan x - sec^2 x and v'(x) = sec x tan x + sec^2 x.
  4. Apply the Quotient Rule Formula: Substitute u(x), v(x), u'(x), and v'(x) into the quotient rule formula: d/dx [u(x) / v(x)] = [v(x)u'(x) - u(x)v'(x)] / (v(x))^2.
  5. Substitute and Expand: Substitute the expressions for u(x), v(x), u'(x), and v'(x) into the formula and expand the products.
  6. Simplify the Expression: Combine like terms and simplify the expression. This may involve algebraic manipulation and trigonometric identities.
  7. Express in Terms of Sine and Cosine (Optional): Rewrite the expression in terms of sine and cosine functions for further simplification or analysis.

Differentiating ln((1 - 3x²) / (1 + 3x²))^(1/2)

Moving on, we now tackle the differentiation of the logarithmic function ln((1 - 3x²) / (1 + 3x²))^(1/2). This problem introduces the concept of the chain rule, a fundamental technique in calculus for differentiating composite functions. A composite function is a function that is formed by applying one function to the result of another function. In this case, we have a logarithmic function composed with a rational function raised to a power. The chain rule states that the derivative of f(g(x)) is f'(g(x)) * g'(x), where f'(x) and g'(x) represent the derivatives of f(x) and g(x), respectively. To effectively apply the chain rule, we must first identify the outer and inner functions. Here, the outer function is the natural logarithm, ln(x), and the inner function is ((1 - 3x²) / (1 + 3x²))^(1/2). The derivative of the outer function, ln(x), is 1/x. To find the derivative of the inner function, we need to apply the power rule and the chain rule again. The power rule states that the derivative of x^n is nx^(n-1). Applying the power rule to ((1 - 3x²) / (1 + 3x²))^(1/2), we get (1/2)((1 - 3x²) / (1 + 3x²))^(-1/2) multiplied by the derivative of the inner function (1 - 3x²) / (1 + 3x²). To differentiate (1 - 3x²) / (1 + 3x²), we need to employ the quotient rule once more. The quotient rule, as discussed earlier, provides a systematic method for differentiating functions expressed as the ratio of two differentiable functions. After applying the quotient rule and simplifying the expression, we obtain the derivative of (1 - 3x²) / (1 + 3x²). Finally, we combine all the derivatives using the chain rule to find the derivative of the original function, ln((1 - 3x²) / (1 + 3x²))^(1/2). The process involves careful substitution, algebraic manipulation, and simplification. Throughout this journey, a keen eye for detail is crucial. Each step, from identifying the outer and inner functions to applying the chain rule and quotient rule, requires careful execution to avoid errors. The final result provides valuable insights into the rate of change of the logarithmic function. It tells us how the function varies as x changes. This knowledge is crucial in various applications, such as analyzing the behavior of logarithmic functions in different contexts. Understanding the chain rule is paramount in calculus, as it allows us to differentiate a wide range of composite functions. Mastering this technique unlocks a deeper understanding of differentiation and its applications. By breaking down the problem into smaller, manageable steps, we can effectively tackle complex differentiation challenges.

Step-by-step Differentiation of ln((1 - 3x²) / (1 + 3x²))^(1/2)

  1. Apply Logarithmic Properties (Optional): Simplify the expression using logarithmic properties. In this case, ln(a^b) = b * ln(a), so the function can be rewritten as (1/2) * ln((1 - 3x²) / (1 + 3x²)).
  2. Identify the Chain Rule: Recognize that the function is a composite function, requiring the chain rule for differentiation.
  3. Define Outer and Inner Functions: Let the outer function be f(u) = (1/2) * ln(u) and the inner function be u(x) = (1 - 3x²) / (1 + 3x²).
  4. Find f'(u) and u'(x): Calculate the derivatives of the outer and inner functions. f'(u) = 1 / (2u). To find u'(x), apply the quotient rule.
  5. Apply the Chain Rule Formula: Substitute f'(u) and u'(x) into the chain rule formula: d/dx [f(u(x))] = f'(u(x)) * u'(x).
  6. Substitute and Simplify: Substitute the expressions for f'(u), u(x), and u'(x) into the formula and simplify the expression. This may involve algebraic manipulation.
  7. Further Simplification (Optional): Simplify the expression further by combining terms and using algebraic identities.

In conclusion, the journey through differentiating trigonometric and logarithmic functions reveals the power and elegance of calculus. By mastering techniques such as the quotient rule and the chain rule, we can unravel the complexities of these functions and gain a deeper understanding of their behavior. Differentiation is not merely a mathematical exercise; it is a tool that empowers us to analyze and model dynamic systems in various fields. From physics to engineering to economics, the principles of calculus provide a framework for understanding change and optimization. The examples presented in this guide serve as a foundation for tackling more complex differentiation problems. Remember that practice is the cornerstone of mastery. The more you engage with differentiation, the more proficient you will become. As you delve deeper into the world of calculus, you will discover its profound connections to other branches of mathematics and its wide-ranging applications in the real world. The ability to differentiate trigonometric and logarithmic functions is a valuable asset for any aspiring mathematician, scientist, or engineer. It opens doors to a world of possibilities, enabling you to solve challenging problems and make meaningful contributions to your chosen field. So, embrace the challenge, persevere through the difficulties, and revel in the satisfaction of mastering this essential skill. The world of calculus awaits, full of fascinating concepts and rewarding discoveries.