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Please help! Calculus: Derivative by limit process!!
4 months ago
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Please help! Calculus: Derivative by limit process!!
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Answer:[tex]\displaystyle f'(x) = 2x + 1[/tex]General Formulas and Concepts:Pre-AlgebraOrder of Operations: BPEMDAS Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right Distributive PropertyAlgebra ITerms/CoefficientsExpandingFactoringFunctionsFunction NotationCalculusLimitsLimit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]Derivatives[tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/tex]Derivative NotationStep-by-step explanation:Step 1: DefineIdentifyf(x) = x² + x - 2Step 2: DifferentiateSubstitute in function [Limit Process]: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{[(x + h)^2 + (x + h) - 2] - (x^2 + x - 2)}{h}[/tex][Brackets] Expand: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{[x^2 + 2hx + h^2 + x + h - 2] - (x^2 + x - 2)}{h}[/tex][Distributive Property] Distribute negative: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{x^2 + 2hx + h^2 + x + h - 2 - x^2 - x + 2}{h}[/tex]Combine like terms: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{2hx + h^2 + h}{h}[/tex]Factor: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{h(2x + h + 1)}{h}[/tex]Simplify: [tex]\displaystyle f'(x) = \lim_{h \to 0} 2x + h + 1[/tex]Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle f'(x) = 2x + 0 + 1[/tex]Simplify: [tex]\displaystyle f'(x) = 2x + 1[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: DerivativesBook: College Calculus 10e