Calculate G′(−3) to 3 significant figures where G(x)=(4x−2)^10(4x2+1x+0)^12

Answer 1

Product rule/chain rule

(4x-2)^10 * 12( 4x^2+x)^11 (8x+1) + (4x^2+x)^12 *10(4x-2)^9 *4 =

simplifies 2nd terms a little bit

12 (4x-2)^10 (4x^2+x)^11 (8x+1) + 40(4x^2+x)^12 (4x-2)^9

factors out common factor quantities

(4x^2+x)^11 (4x-2)^9 [ 12(4x-2)(8x+1) + 40(4x^2+x) ]

FOIL

(4x^2+x)^11 (4x-2)^9 [ 12(32x^2-12x-2) + 40(4x^2+x)]

(4x^2+x)^11 (4x-2)^9 [ 384x^2 - 144x - 24 + 160x^2 + 40x]

(4x^2+x)^11 (4x-2)^9 [ 544x^2 - 104x - 24]

Into that you must plug x=-3:

G'(-3) = (33^11) (-14)^9 [ 544*9 + 312 - 24]

it will be a severely negative number