Question

найти у'-? а)2х^5*е^8х б)sin5x-4cos2x в)2en(5x+3) г)4х^3:2x+1​

Answer 1

$1)\ \ (2x^5\cdot e^{8x})'=10x^4\cdot e^{8x}+2x^5\cdot 8e^{8x}\\\\2)\ \ (sin5x-4cos2x)'=5cos5x+8sin2x\\\\3)\ \ (2e^{5x+3})'=10e^{5x+3}\\\\4)\ \ \Big(\dfrac{4x^3}{2x+1}\Big)'=\dfrac{12x^2\, (2x+1)-8x^3}{(2x+1)'}=\dfrac{16x^3+12x^2}{(2x+1)^2}=\dfrac{4x^2\, (4x+3)}{(2x+1)^2}$