Question

решите 2 пример ктонибудь

Answer 1

$y'=( \frac{e^{-x^2}}{(1-3x)^4})'= \frac{(e^{-x^2})'*(1-3x)^4-e^{-x^2}*((1-3x)^4)'}{(1-3x)^8}=$$=\frac{(-x^2)'e^{-x^2}*(1-3x)^4-e^{-x^2}*4(1-3x)^3(1-3x)'}{(1-3x)^8}=$$\frac{-2xe^{-x^2}*(1-3x)^4-e^{-x^2}*4(1-3x)^3(-3)}{(1-3x)^8}=\frac{-2xe^{-x^2}*(1-3x)^4+12e^{-x^2}(1-3x)^3}{(1-3x)^8}=$$\frac{12e^{-x^2}+(6x^2-2x)e^{-x^2}}{(1-3x)^5}=\frac{e^{-x^2}(6x^2-2x+12)}{(1-3x)^5}$