Find the indicated derivative.

Answer:[tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{(9t - 8)^5(45t + 116)}{(t + 2)^2}[/tex]General Formulas and Concepts:CalculusDifferentiationDerivativesDerivative NotationDerivative 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:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]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: DefineIdentify[tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg)[/tex]Step 2: DifferentiateDerivative Rule [Quotient Rule]:                                                                   [tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{[(9t - 8)^6]'(t + 2) - (9t - 8)^6(t + 2)'}{(t + 2)^2}[/tex]Basic Power Rule [Chain Rule, Multiplied Constant]:                                 [tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{6(9t - 8)^5(9t - 8)'(t + 2) - (9t - 8)^6}{(t + 2)^2}[/tex]Basic Power Rule [Multiplied Constant, Addition/Subtraction]:                 [tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{54(9t - 8)^5(t + 2) - (9t - 8)^6}{(t + 2)^2}[/tex]Factor:                                                                                                           [tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{(9t - 8)^5 \big[ 54(t + 2) - (9t - 8) \big] }{(t + 2)^2}[/tex]Expand:                                                                                                         [tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{(9t - 8)^5 \big[ 54t + 108 - 9t + 8 \big] }{(t + 2)^2}[/tex]Combine like terms:                                                                                     [tex]\displaystyle \frac{d}{dt} \bigg( \frac{(9t - 8)^6}{t + 2} \bigg) = \frac{(9t - 8)^5(45t + 116)}{(t + 2)^2}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: Differentiation