Question

Найдите производную
((2х+3)*(4x^2-3x+7))

Answer 1

$((2x+3)*(4x^2-3x+7))'=24x^2+12x+5\\Variantquad1\((2x+3)*(4x^2-3x+7))'=(2x+3)'(4x^2-3x+7)+(2x+3)(4x^2-3x+7)'=\=8x^2-6x+14+(2x+3)(8x-3)=8x^2-6x+14+16x^2+18x-9=\=24x^2+12x+5\\Variantquad2\((2x+3)*(4x^2-3x+7))'=(8x^3-6x^2+14x+12x^2-9x+21)'=\=(8x^3+6x^2+5x-21)'=24x^2+12x+5$