Respuesta :

LammettHash

Since [tex]\dfrac{\mathrm d}{\mathrm dx}\arcsin x=\dfrac1{\sqrt{1-x^2}}[/tex], you have

[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}2\arcsin(x-1)=2\frac{\mathrm d}{\mathrm dx}\arcsin(x-1)=2\frac{\dfrac{\mathrm d}{\mathrm dx}(x-1)}{\sqrt{1-(x-1)^2}}=\frac2{\sqrt{2x-x^2}}[/tex]

Space

Answer:

[tex]\displaystyle f'(x) = \frac{2}{\sqrt{1 - (x - 1)^2}}[/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]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle f(x) = 2 \arcsin (x - 1)[/tex]

Step 2: Differentiate

  1. Derivative Property [Multiplied Constant]:                                                   [tex]\displaystyle f'(x) = 2 \frac{d}{dx}[\arcsin (x - 1)][/tex]
  2. Trigonometric Differentiation [Derivative Rule - Chain Rule]:                   [tex]\displaystyle f'(x) = 2 \bigg( \frac{(x - 1)'}{\sqrt{1 - (x - 1)^2}} \bigg)[/tex]
  3. Basic Power Rule [Derivative Properties]:                                                   [tex]\displaystyle f'(x) = 2 \bigg( \frac{1}{\sqrt{1 - (x - 1)^2}} \bigg)[/tex]
  4. Simplify:                                                                                                         [tex]\displaystyle f'(x) = \frac{2}{\sqrt{1 - (x - 1)^2}}[/tex]

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

Unit: Differentiation

Otras preguntas