Derivative Of Sinx 1 Cosx

Unveiling the Mystery: Deriving the Derivative of sin x / (1 - cos x)

Understanding the derivative of a function is fundamental in calculus. This article delves into the derivation of the derivative of the trigonometric function sin x / (1 - cos x), exploring various techniques and providing a detailed, step-by-step explanation. We’ll not only find the derivative but also explore the underlying mathematical principles and address common questions, making this a comprehensive guide for students and anyone interested in deepening their calculus knowledge.

Introduction

The function sin x / (1 - cos x) presents a unique challenge due to its combination of trigonometric functions and a denominator that involves subtraction. Finding its derivative requires a robust understanding of differentiation rules, specifically the quotient rule and trigonometric identities. This article aims to illuminate the process, breaking down the derivation into manageable steps and providing insightful explanations along the way. The keyword, "derivative of sinx 1 cosx," and related semantic keywords will naturally appear throughout the text to ensure organic SEO optimization.

Understanding the Necessary Tools: Differentiation Rules and Trigonometric Identities

Before diving into the derivation, let's review the essential mathematical tools we’ll need:

• Quotient Rule: The quotient rule is crucial for differentiating functions in the form of f(x) / g(x). The rule states: d/dx [f(x) / g(x)] = [g(x)f'(x) - f(x)g'(x)] / [g(x)]².

• Derivatives of Trigonometric Functions: We need to know the derivatives of sin x and cos x:

  • d/dx (sin x) = cos x
  • d/dx (cos x) = -sin x

• Trigonometric Identities: Several trigonometric identities will simplify the process. Key identities include:

  • sin²x + cos²x = 1
  • 1 - cos²x = sin²x
  • sin 2x = 2sin x cos x

Step-by-Step Derivation of the Derivative

Let's denote our function as y = sin x / (1 - cos x). We will apply the quotient rule to find its derivative, dy/dx.

1. Identifying f(x) and g(x): In our function, f(x) = sin x and g(x) = (1 - cos x).

2. Finding the Derivatives of f(x) and g(x):

   • f'(x) = d/dx (sin x) = cos x
   • g'(x) = d/dx (1 - cos x) = sin x

3. Applying the Quotient Rule: Now, we substitute f(x), g(x), f'(x), and g'(x) into the quotient rule formula:

   dy/dx = [(1 - cos x)(cos x) - (sin x)(sin x)] / (1 - cos x)²

4. Simplifying the Expression: Let's expand and simplify the numerator:

   dy/dx = [cos x - cos²x - sin²x] / (1 - cos x)²

5. Using Trigonometric Identities: Recall the identity sin²x + cos²x = 1. We can substitute this into our expression:

   dy/dx = [cos x - (sin²x + cos²x)] / (1 - cos x)² dy/dx = [cos x - 1] / (1 - cos x)²

6. Further Simplification: Notice that the numerator is simply the negative of the denominator. Therefore:

   dy/dx = -1 / (1 - cos x)

Alternative Approach Using the Chain Rule and Trigonometric Substitution

An alternative approach involves employing a clever trigonometric substitution. This method showcases the versatility of calculus techniques.

1. Half-Angle Formula: Recall the half-angle formula for tan(x/2) which is: tan(x/2) = sin x / (1 + cos x)

2. Manipulating the original function: This is crucial. Let's multiply our original function by (1 + cos x)/(1 + cos x), which is essentially multiplying by 1, and doesn’t change the value. This gives us:

   y = [sin x (1 + cos x)] / [(1 - cos x)(1 + cos x)] y = [sin x (1 + cos x)] / (1 - cos²x)

3. Using Trigonometric Identities: Using the identity 1 - cos²x = sin²x, we get:

   y = [sin x (1 + cos x)] / sin²x y = (1 + cos x) / sin x

4. Simplifying and rewriting the function: This gives us:

   y = csc x + cot x

5. Differentiating using known derivatives: Now, we can use the known derivatives of cosecant (csc) and cotangent (cot):

   dy/dx = d/dx (csc x + cot x) = -csc x cot x - csc²x

6. Comparing the results: Both methods lead to different yet equivalent forms of the derivative. They are equivalent expressions, despite their different appearances. It’s a testament to the richness of mathematics that a problem can be solved through multiple routes!

Explanation of the Results

Both methods yielded different forms of the derivative, highlighting the flexibility of mathematical approaches. The first method yielded a more compact form, -1 / (1 - cos x), suitable for direct application. The second method, using the half-angle substitution, resulted in an expression involving cosecant and cotangent. Both forms are mathematically correct and can be used depending on the context and preferred style. It's important to understand that equivalent functions can sometimes appear quite different.

Common Questions and Clarifications (FAQ)

• Why is the quotient rule necessary? The quotient rule is essential because our function involves a numerator and a denominator. Direct differentiation of the numerator and denominator separately wouldn't yield the correct derivative.

• Can I use L'Hopital's rule? L'Hopital's rule is used to evaluate indeterminate forms (like 0/0 or ∞/∞). While our function might appear to have such a form at certain points, it's generally more efficient and straightforward to use the quotient rule directly.

• Are both derivative forms equivalent? Yes. They represent the same function, just in different forms. One may be more convenient to use than the other in various calculations or applications.

• What are the applications of this derivative? The derivative of sin x / (1 - cos x) finds applications in various fields, including physics (related rate problems involving oscillatory motion) and engineering (modeling certain types of curves).

Conclusion

Deriving the derivative of sin x / (1 - cos x) requires a solid grasp of differentiation rules and trigonometric identities. This article presented a detailed step-by-step solution using the quotient rule, complemented by an alternative approach using trigonometric substitution. Understanding both methods provides valuable insight into the richness of calculus and the flexibility of its techniques. The exploration of different approaches reinforces the idea that there isn't always just one way to solve a problem, and that each method can offer different advantages depending on the context. The derived expressions, despite their distinct forms, are mathematically equivalent and can be applied to solve various problems in calculus and beyond. Remember to practice applying these techniques to solidify your understanding.