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Fundamental Properties of Differentiation

Michael Green

Michael Green

6 min read

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Study Guide Overview

This study guide covers derivatives of tangent and reciprocal trigonometric functions. It includes the derivative of tan(x) and tan(kx) using the quotient and chain rules. It also defines and derives the derivatives of sec(x), csc(x), and cot(x). The guide provides worked examples, practice questions, a glossary, and exam strategies.

Study Notes: Derivatives of Trigonometric Functions

Table of Contents

  1. Derivative of the Tangent Function
  2. Derivatives of Reciprocal Trigonometric Functions
  3. Practice Questions
  4. Glossary
  5. Summary and Key Takeaways
  6. Exam Strategy

Derivative of the Tangent Function

What is the Derivative of tan⁡x\tan x?

Key Concept

If f(x)=tan⁡xf(x) = \tan x, then the derivative is given by: f′(x)=sec⁡2xf'(x) = \sec^2 x

This can be shown using the identity tan⁡x≡sin⁡xcos⁡x\tan x \equiv \frac{\sin x}{\cos x} and the quotient rule.

The quotient rule states that if y=uvy = \frac{u}{v}, then y′=u′v−uv′v2y' = \frac{u'v - uv'}{v^2}

Let u=sin⁡xu = \sin x and v=cos⁡xv = \cos x. - u′=cos⁡xu' = \cos x - v′=−sin⁡xv' = -\sin x

Applying the quotient rule: y′=cos⁡x⋅cos⁡x+sin⁡x⋅sin⁡xcos⁡2xy' = \frac{\cos x \cdot \cos x + \sin x \cdot \sin x}{\cos^2 x} Simplifying: y′=cos⁡2x+sin⁡2xcos⁡2xy' = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} Using the identity sin⁡2x+cos⁡2x≡1\sin^2 x + \cos^2 x \equiv 1: y′=1cos⁡2x=sec⁡2xy' = \frac{1}{\cos^2 x} = \sec^2 x

What is the Derivative of tan⁡kx\tan kx?

Key Concept

If f(x)=tan⁡kxf(x) = \tan kx, then the derivative is: f′(x)=ksec⁡2kxf'(x) = k \sec^2 kx

This is a result of applying the chain rule. It can also be shown using the quotient rule for sin⁡kxcos⁡kx\frac{\sin kx}{\cos kx}.

Worked Examples

Differentiate the following functions:

(a) f(x)=tan⁡3xf(x) = \tan 3x

Answer:

Using the rule for tan⁡kx\tan kx: f′(x)=3sec⁡23xf'(x) = 3 \sec^2 3x

(b) g(x)=3x2tan⁡xg(x) = 3x^2 \tan x

Answer:

This is a product of two terms, so use the product rule: y′=u′v+uv′y' = u'v + uv'

Let u=3x2u = 3x^2 and v=tan⁡xv = \tan x.

  • u′=6xu' = 6x
  • v′=sec⁡2xv' = \sec^2 x

Applying the product rule: y′=6x⋅tan⁡x+3x2⋅sec⁡2x=3x(2tan⁡x+xsec⁡2x)y' = 6x \cdot \tan x + 3x^2 \cdot \sec^2 x = 3x (2 \tan x + x \sec^2 x)

Derivatives of Reciprocal Trigonometric Functions

What are the Reciprocal Trig Functions?

Key Concept

The reciprocal trigonometric functions are:

  • csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}
  • sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}
  • cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}
**csc** is sometimes written as **cosec**. You can remember the functions by looking at the **third letter** of each: - co**s**ec is the reciprocal of **s**in - se**c** is the reciprocal of **c**os - co**t** is the reciprocal of **t**an

What are the Derivatives of the Reciprocal Trig Functions?

Key Concept

If f(x)=csc⁡xf(x) = \csc x, then: f′(x)=−cot⁡xcsc⁡xf'(x) = -\cot x \csc x

If g(x)=sec⁡xg(x) = \sec x, then: g′(x)=tan⁡xsec⁡xg'(x) = \tan x \sec x

If h(x)=cot⁡xh(x) = \cot x, then: h′(x)=−csc⁡2xh'(x) = -\csc^2 x

These results can be remembered or derived using the reciprocal trig function definitions and the quotient rule.

How to Derive the Derivatives

Derivative of csc⁡x\csc x

Recall that csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}. Apply the quotient rule: y′=u′v−uv′v2y' = \frac{u'v - uv'}{v^2}

Let u=1u = 1 and v=sin⁡xv = \sin x. - u′=0u' = 0 - v′=cos⁡xv' = \cos x

Applying the quotient rule: y′=0−cos⁡x(sin⁡x)2=−cos⁡xsin⁡2xy' = \frac{0 - \cos x}{(\sin x)^2} = \frac{-\cos x}{\sin^2 x}

Simplify using the identities csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x} and cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}: y′=−cot⁡xcsc⁡xy' = -\cot x \csc x

Derivative of sec⁡x\sec x

Recall that sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}. Apply the quotient rule: y′=u′v−uv′v2y' = \frac{u'v - uv'}{v^2}

Let u=1u = 1 and v=cos⁡xv = \cos x. - u′=0u' = 0 - v′=−sin⁡xv' = -\sin x

Applying the quotient rule: y′=0−(−sin⁡x)(cos⁡x)2=sin⁡xcos⁡2xy' = \frac{0 - (-\sin x)}{(\cos x)^2} = \frac{\sin x}{\cos^2 x}

Simplify using the identity sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}: y′=tan⁡xsec⁡xy' = \tan x \sec x

Derivative of cot⁡x\cot x

Recall that cot⁡x=1tan⁡x=cos⁡xsin⁡x\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}. Apply the quotient rule: y′=u′v−uv′v2y' = \frac{u'v - uv'}{v^2}

Let u=cos⁡xu = \cos x and v=sin⁡xv = \sin x. - u′=−sin⁡xu' = -\sin x - v′=cos⁡xv' = \cos x

Applying the quotient rule: y′=−sin⁡x⋅sin⁡x−cos⁡x⋅cos⁡x(sin⁡x)2=−sin⁡2x−cos⁡2xsin⁡2xy' = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}

Simplify using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1: y′=−1sin⁡2x=−csc⁡2xy' = \frac{-1}{\sin^2 x} = -\csc^2 x

Worked Example

Show that the derivative of f(x)=csc⁡xsec⁡xf(x) = \csc x \sec x is f′(x)=sec⁡2x−csc⁡2xf'(x) = \sec^2 x - \csc^2 x.

This is a product of two functions, so use the product rule: y′=u′v+uv′y' = u'v + uv'

Let u=csc⁡xu = \csc x and v=sec⁡xv = \sec x. Differentiate using the known results:

  • u′=−cot⁡xcsc⁡xu' = -\cot x \csc x
  • v′=tan⁡xsec⁡xv' = \tan x \sec x

Apply the product rule: y′=−cot⁡xcsc⁡x⋅sec⁡x+csc⁡x⋅tan⁡xsec⁡xy' = -\cot x \csc x \cdot \sec x + \csc x \cdot \tan x \sec x

Using trigonometric identities to simplify: y′=(−cos⁡xsin⁡x⋅1sin⁡x⋅1cos⁡x)+(1sin⁡x⋅sin⁡xcos⁡x⋅1cos⁡x)y' = \left( \frac{-\cos x}{\sin x} \cdot \frac{1}{\sin x} \cdot \frac{1}{\cos x} \right) + \left( \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \right)

Simplify further: y′=−1sin⁡2x+1cos⁡2x=−csc⁡2x+sec⁡2xy' = \frac{-1}{\sin^2 x} + \frac{1}{\cos^2 x} = -\csc^2 x + \sec^2 x

Practice Questions

Practice Question
  1. Differentiate f(x)=tan⁡5xf(x) = \tan 5x.
  2. Differentiate g(x)=x2sec⁡xg(x) = x^2 \sec x.
  3. Show that the derivative of h(x)=cot⁡2xh(x) = \cot 2x is h′(x)=−2csc⁡22xh'(x) = -2 \csc^2 2x.
  4. Differentiate k(x)=2xcsc⁡xk(x) = 2x \csc x.

Glossary

  • Quotient Rule: A rule for differentiating the quotient of two functions.
  • Chain Rule: A rule for differentiating compositions of functions.
  • Secant (sec⁡\sec): Reciprocal of cosine, sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.
  • Cosecant (csc⁡\csc): Reciprocal of sine, csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}.
  • Cotangent (cot⁡\cot): Reciprocal of tangent, cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}.

Summary and Key Takeaways

  • The derivative of tan⁡x\tan x is sec⁡2x\sec^2 x.
  • The derivative of tan⁡kx\tan kx is ksec⁡2kxk \sec^2 kx.
  • The derivatives of reciprocal trig functions can be derived using the quotient rule.
  • Memorizing the derivatives of sec⁡x\sec x, csc⁡x\csc x, and cot⁡x\cot x is useful for solving problems quickly.

Exam Strategy

Exam Tip
  • Always start by identifying which differentiation rule applies: product rule, quotient rule, or chain rule.
  • Simplify trigonometric expressions using identities before differentiating.
  • Practice deriving the formulas for derivatives of reciprocal trigonometric functions to reinforce understanding.

By following these notes and practicing the provided questions, you'll be well-prepared for questions involving the derivatives of trigonometric functions on your exam.

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Question 1 of 11

What is the derivative of f(x)=tan⁡4xf(x) = \tan 4x ? 🚀

4sec⁡24x4 \sec^2 4x

sec⁡24x\sec^2 4x

4tan⁡4xsec⁡4x4 \tan 4x \sec 4x

4cot⁡4x4 \cot 4x