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If f(x) = 9x10 tan−1x, find f '(x).

90x9tan−1x + 9x10 1 over the quantity 1 plus x squared
90x9tan−1x + 9x10tan−2x
90x9tan−1x − 9x10 1 over the square root of the quantity 1 minus x squared
90x9tan−1x + 9x10 1 over the square root of the quantity 1 minus x squared
those are my choices

Sagot :

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Answer:

[tex]\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                             [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle f(x) = 9x^{10} \tan^{-1}(x)[/tex]

Step 2: Differentiate

  1. [Function] Derivative Rule [Product Rule]:                                                   [tex]\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)][/tex]
  2. Rewrite [Derivative Property - Multiplied Constant]:                                  [tex]\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)][/tex]
  3. Basic Power Rule:                                                                                         [tex]\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)][/tex]
  4. Arctrig Derivative:                                                                                         [tex]\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}[/tex]

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

Unit: Differentiation

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