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what is the derivative of secx​

Sagot :

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Answer:

[tex]\displaystyle \frac{d}{dx}[\sec x] = \sec x \tan x[/tex]

General Formulas and Concepts:

Pre-Calculus

  • Trigonometric Identities

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

Step-by-step explanation:

*Note:

This is a known trigonometric derivative.

Step 1: Define

Identify

[tex]\displaystyle y = \sec x[/tex]

Step 2: Differentiate

  1. Rewrite [Trigonometric Identities]:                                                               [tex]\displaystyle y = \frac{1}{\cos x}[/tex]
  2. Derivative Rule [Quotient Rule]:                                                                   [tex]\displaystyle y' = \frac{(1)' \cos x - 1(\cos x)'}{\cos^2 x}[/tex]
  3. Basic Power Rule:                                                                                         [tex]\displaystyle y' = \frac{(0) \cos x - 1(\cos x)'}{\cos^2 x}[/tex]
  4. Trigonometric Differentiation:                                                                       [tex]\displaystyle y' = \frac{(0) \cos x + 1(\sin x)}{\cos^2 x}[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle y' = \frac{\sin x}{\cos^2 x}[/tex]
  6. Rewrite:                                                                                                         [tex]\displaystyle y' = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}[/tex]
  7. Rewrite [Trigonometric Identities]:                                                               [tex]\displaystyle y' = \tan x \sec x[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

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