AP Calculus BC 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions Study Notes - New Syllabus
AP Calculus BC 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions Study Notes- New syllabus
AP Calculus BC 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Recognizing opportunities to apply derivative rules can simplify differentiation.
Key Concepts:
- Derivatives of Tangent, Cotangent, Secant, and Cosecant Functions
Derivatives of Tangent, Cotangent, Secant, and Cosecant Functions
Derivative of Tangent Function
The tangent function is defined as \( \tan x = \dfrac{\sin x}{\cos x} \). To differentiate \( \tan x \), we can apply the Quotient Rule. However, the result is well known and can be directly stated:
Formula: If \( f(x) = \tan x \), then \( f'(x) = \sec^2 x \)
This derivative is valid for all values of \( x \) where \( \cos x \neq 0 \), since \( \tan x \) is undefined when \( \cos x = 0 \).
Example:
Differentiate \( f(x) = \tan x \)
▶️Answer/Explanation
\( f'(x) = \sec^2 x \)
Example:
Differentiate \( f(x) = 5\tan x \)
▶️Answer/Explanation
Using the constant multiple rule: \( f'(x) = 5 \cdot \frac{d}{dx}(\tan x) = 5\sec^2 x \)
Derivative of Cotangent Function
The cotangent function is defined as \( \cot x = \dfrac{\cos x}{\sin x} \). To differentiate \( \cot x \), we could use the Quotient Rule. However, the standard result is:
Formula:
If \( f(x) = \cot x \), then \( f'(x) = -\csc^2 x \)
This derivative is valid for all \( x \) where \( \sin x \neq 0 \), since \( \cot x \) is undefined when \( \sin x = 0 \).
Example:
Differentiate \( f(x) = \cot x \)
▶️Answer/Explanation
\( f'(x) = -\csc^2 x \)
Example:
Differentiate \( f(x) = -2\cot x \)
▶️Answer/Explanation
Using the constant multiple rule:
\( f'(x) = -2 \cdot \frac{d}{dx}(\cot x) = -2(-\csc^2 x) = 2\csc^2 x \)
Derivative of Secant Function
The secant function is defined as \( \sec x = \dfrac{1}{\cos x} \). Its derivative is a standard result:
Formula:
If \( f(x) = \sec x \), then \( f'(x) = \sec x \tan x \)
This result is derived using the chain rule and the identity of \( \sec x \) as a reciprocal of \( \cos x \).
Domain Note: The derivative exists for all \( x \) where \( \cos x \neq 0 \), since \( \sec x \) is undefined where \( \cos x = 0 \).
Example:
Differentiate \( f(x) = \sec x \)
▶️Answer/Explanation
Using the standard formula:
\( f'(x) = \sec x \tan x \)
Example:
Differentiate \( f(x) = 3\sec x \)
▶️Answer/Explanation
Apply the constant multiple rule:
\( f'(x) = 3 \cdot \frac{d}{dx}(\sec x) = 3\sec x \tan x \)
Derivative of Cosecant Function
The cosecant function is defined as \( \csc x = \dfrac{1}{\sin x} \). Its derivative is another standard result in calculus:
Formula:
If \( f(x) = \csc x \), then \( f'(x) = -\csc x \cot x \)
This formula can be derived using the chain rule and the identity of \( \csc x \) as the reciprocal of \( \sin x \).
Domain Note: The derivative exists wherever \( \sin x \neq 0 \), since \( \csc x \) is undefined where \( \sin x = 0 \).
Example:
Differentiate \( f(x) = \csc x \)
▶️Answer/Explanation
Using the standard formula:
\( f'(x) = -\csc x \cot x \)
Example:
Differentiate \( f(x) = 2\csc x \)
▶️Answer/Explanation
Apply the constant multiple rule:
\( f'(x) = 2 \cdot \frac{d}{dx}(\csc x) = -2\csc x \cot x \)
| Function | Derivative |
|---|---|
| \( \tan x \) | \( \sec^2 x \) |
| \( \cot x \) | \( -\csc^2 x \) |
| \( \sec x \) | \( \sec x \tan x \) |
| \( \csc x \) | \( -\csc x \cot x \) |