Derivative Of Cotangent (cot X): Formula And Proof

by Wholesomestory Johnson 51 views

Hello there! Let's dive into the world of calculus and explore the derivative of cotangent, often represented as cot x. I'll break down the formula, and provide a step-by-step proof using the first principle, so you understand it completely. Get ready for a clear, detailed, and correct explanation!

Correct Answer

The derivative of cot x is -csc²x, which can also be written as -1/sin²x.

Detailed Explanation

Alright, let’s unravel this step-by-step. Understanding the derivative of cot x is crucial in many areas of calculus and its applications. We'll explore the formula and derive it using the first principle.

What is a Derivative?

Before we begin, let's quickly recap what a derivative is. In simple terms, the derivative of a function at a point represents the instantaneous rate of change of the function at that point. Geometrically, it’s the slope of the tangent line to the function's graph at that point.

The Formula

The formula for the derivative of cot x is:

d/dx (cot x) = -csc²x

Where:

  • d/dx is the derivative operator.
  • cot x is the cotangent function.
  • csc x is the cosecant function, and csc x = 1/sin x.

Proof by First Principle

Now, let's prove this using the first principle (also known as the definition of the derivative). The first principle defines the derivative of a function f(x) as:

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

For cot x, this becomes:

d/dx (cot x) = lim (h→0) [cot(x + h) - cot(x)] / h

Let's break this down step by step:

  1. Rewrite cot in terms of sin and cos

    We know that cot x = cos x / sin x. So, we can rewrite our expression as:

    lim (h→0) [cos(x + h) / sin(x + h) - cos(x) / sin(x)] / h

  2. Combine the Fractions

    To combine the fractions, we need a common denominator:

    lim (h→0) [cos(x + h)sin(x) - cos(x)sin(x + h)] / [h * sin(x + h)sin(x)]

  3. Apply the Trigonometric Identity for sin(A - B)

    Notice that the numerator resembles the sine difference formula: sin(A - B) = sin(A)cos(B) - cos(A)sin(B). We can rewrite the numerator using the identity sin(A - B) = sinAcosB - cosAsinB. It can be converted to a difference between two sin functions, as cos(x + h)sin(x) - cos(x)sin(x + h) = -[sin(x + h - x)] = sin(x - (x+h))

    The numerator looks similar to the sine difference identity. Recall the sine subtraction formula: sin(A - B) = sinAcosB - cosAsinB. So, we can rearrange the numerator as follows, by factoring out a negative sign:

    lim (h→0) - [sin(x + h - x)] / [h * sin(x + h)sin(x)] lim (h→0) - [sin(h)] / [h * sin(x + h)sin(x)]

  4. Separate the Limit

    We can separate the limit into parts. Let's rewrite the limit as:

    lim (h→0) - [sin(h) / h] * [1 / (sin(x + h)sin(x))]

  5. Evaluate the Limits

    We know a fundamental limit: lim (h→0) sin(h) / h = 1.

    So, we have:

    -1 * [1 / (sin(x + 0)sin(x))]

    Which simplifies to:

    -1 / sin²(x)

  6. Relate to Cosecant

    Finally, since csc x = 1 / sin x, we have:

    d/dx (cot x) = -csc²x

And there you have it – the derivative of cot x derived using the first principle!

Alternative Method: Using Quotient Rule

Another way to derive the derivative of cot x is by using the quotient rule.

  1. Rewrite cot x

    cot x = cos x / sin x

  2. Apply the Quotient Rule

    The quotient rule states that if you have a function f(x) = u(x) / v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]²

    Here, u(x) = cos x and v(x) = sin x.

    u'(x) = -sin x v'(x) = cos x

    So:

    d/dx (cot x) = [(-sin x)(sin x) - (cos x)(cos x)] / (sin²x)

  3. Simplify

    d/dx (cot x) = [-sin²x - cos²x] / sin²x d/dx (cot x) = -[sin²x + cos²x] / sin²x

  4. Use the Pythagorean Identity

    Recall the fundamental trigonometric identity: sin²x + cos²x = 1.

    d/dx (cot x) = -1 / sin²x

  5. Relate to Cosecant

    d/dx (cot x) = -csc²x

This confirms the formula using the quotient rule as well.

Tips for Memorization

  • Relate to Sine and Cosine: Remember that cot x is essentially cos x over sin x. This can help you recall the quotient rule derivation. Alternatively, understand how cot x is derived and use the first principle to understand its behavior.
  • Cosecant Connection: The derivative involves csc x. Remembering that csc x = 1/sin x can help. Notice the negative sign comes into play when you differentiate. You can keep track by remembering the negative derivatives start with 'c' - cos, cot, csc.
  • Practice Problems: Solve plenty of exercises to reinforce your understanding. Practice makes perfect, especially with calculus!

Real-World Applications

The derivative of cotangent has applications in various fields:

  • Physics: In the analysis of oscillatory systems and wave phenomena.
  • Engineering: In signal processing and control systems.
  • Mathematics: In the study of trigonometric functions and calculus.

Key Takeaways

  • The derivative of cot x is -csc²x.
  • You can derive this formula using the first principle or the quotient rule.
  • Understanding derivatives is crucial for analyzing rates of change.
  • Practice regularly to master the concept.
  • Applications span physics, engineering, and more.