Advanced Placement (AP), 11.12.2019 10:31, dele53

Ap calculus ab. i'm completely lost

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answered: DESIREEHANCOTTE10

$\displaystyle f'(0) = \frac{4}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

Terms/CoefficientsFactoringFunctionsFunction NotationExponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

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

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

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

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

Explanation:

Step 1: Define

Identify

$\displaystyle f(x) = (x^2 - 2x - 1)^\bigg{\frac{2}{3}}$

Step 2: Differentiate

Chain Rule: $\displaystyle f'(x) = \frac{d}{dx}[(x^2 - 2x - 1)^\bigg{\frac{2}{3}}] \cdot \frac{d}{dx}[(x^2 - 2x - 1)]$ Basic Power Rule {Derivative Property - Subtraction]: $\displaystyle f'(x) = \frac{2}{3}(x^2 - 2x - 1)^\bigg{\frac{2}{3} - 1} \cdot (2x^{2 - 1} - 2x^{1 - 1} - 0)$ Simplify: $\displaystyle f'(x) = \frac{2}{3}(x^2 - 2x - 1)^\bigg{\frac{-1}{3}} \cdot (2x - 2)$ Rewrite [Exponential Rule - Rewrite]: $\displaystyle f'(x) = \frac{2(2x - 2)}{3(x^2 - 2x - 1)^\bigg{\frac{1}{3}}}$ Factor: $\displaystyle f'(x) = \frac{4(x - 1)}{3(x^2 - 2x - 1)^\bigg{\frac{1}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle f'(0) = \frac{4(0 - 1)}{3[0^2 - 2(0) - 1]^\bigg{\frac{1}{3}}}$ [Order of Operations] Simplify: $\displaystyle f'(0) = \frac{4}{3}$

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

Unit: Derivatives

Book: College Calculus 10e

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answered: Guest

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