Math Problem Statement

Show that the derivative of cot(x) is -csc^2(x).

Solution

To show that:

ddx(cotx)=csc2x\frac{d}{dx} (\cot x) = -\csc^2 x

Step-by-step differentiation:

  1. Express cotx\cot x: cotx=cosxsinx\cot x = \frac{\cos x}{\sin x} Now we need to differentiate this using the quotient rule.

  2. Quotient Rule: If f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, then: ddx(g(x)h(x))=h(x)g(x)g(x)h(x)[h(x)]2\frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{h(x) g'(x) - g(x) h'(x)}{[h(x)]^2} Here, g(x)=cosxg(x) = \cos x and h(x)=sinxh(x) = \sin x.

  3. Differentiate cosx\cos x and sinx\sin x: ddx(cosx)=sinx,ddx(sinx)=cosx\frac{d}{dx} (\cos x) = -\sin x, \quad \frac{d}{dx} (\sin x) = \cos x

  4. Apply the quotient rule: ddx(cotx)=sinx(sinx)cosxcosx(sinx)2\frac{d}{dx} (\cot x) = \frac{\sin x \cdot (-\sin x) - \cos x \cdot \cos x}{(\sin x)^2} Simplifying the numerator: =sin2xcos2xsin2x= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}

  5. Use the Pythagorean identity: sin2x+cos2x=1\sin^2 x + \cos^2 x = 1 Therefore, the expression becomes: =1sin2x= \frac{-1}{\sin^2 x}

  6. Simplify: 1sin2x=csc2x\frac{-1}{\sin^2 x} = -\csc^2 x

Thus, we have shown that:

ddx(cotx)=csc2x\frac{d}{dx} (\cot x) = -\csc^2 x

Would you like more details on any part of this solution?

Related questions:

  1. How do you differentiate other trigonometric functions like secx\sec x or cscx\csc x?
  2. What is the derivative of tanx\tan x, and how does it compare to cotx\cot x?
  3. Can you apply the quotient rule to non-trigonometric functions?
  4. How does the chain rule apply to trigonometric functions?
  5. What are the integration counterparts of these trigonometric derivatives?

Tip:

Memorizing the derivatives of basic trigonometric functions will save you a lot of time in calculus problems.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Derivatives

Formulas

Quotient Rule
Derivative of sin(x)
Derivative of cos(x)
Pythagorean Identity

Theorems

Quotient Rule
Pythagorean Identity: sin^2(x) + cos^2(x) = 1

Suitable Grade Level

Grades 11-12 (A-level Calculus)