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dmitriy555 [2]
3 years ago
9

Derivative by first principle logsecx²​

Mathematics
1 answer:
Maru [420]3 years ago
5 0

Answer:

\displaystyle \frac{d}{dx}[\log \big( \sec (x^2) \big)] = \frac{2x \tan x^2}{\ln 10}

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                 \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = \log \big( \sec (x^2) \big)

<u>Step 2: Differentiate</u>

  1. Logarithmic Differentiation [Derivative Rule - Chain Rule]:                      \displaystyle y' = \frac{(\sec x^2)'}{\ln (10) \sec x^2}
  2. Trigonometric Differentiation [Derivative Rule - Chain Rule]:                   \displaystyle y' = \frac{\sec x^2 \tan x^2 (x^2)'}{\ln (10) \sec x^2}
  3. Simplify:                                                                                                         \displaystyle y' = \frac{\tan x^2 (x^2)'}{\ln 10}
  4. Basic Power Rule:                                                                                        \displaystyle y' = \frac{2x \tan x^2}{\ln 10}

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

Unit: Differentiation

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