Задача 76967 W \ \ 23›_ | й \_ eI /\ 6 A 157 \\ \"'...
Условие
Решение
[m]y' = \frac{(2x^3)'(x^2-4) - 2x^3(x^2-4)'}{(x^2-4)^2} = \frac{6x^2(x^2-4) - 2x^3 \cdot 2x}{(x^2-4)^2} = \frac{6x^4-24x^2 - 4x^4}{(x^2-4)^2} = \frac{2x^4-24x^2}{(x^2-4)^2}[/m]
[m]y'' = \frac{(2x^4-24x^2)'(x^2-4)^2 - (2x^4-24x^2)((x^2-4)^2)'}{(x^2-4)^4} = \frac{(8x^3-48x)(x^2-4)^2 - (2x^4-24x^2) \cdot 2(x^2-4) \cdot 2x}{(x^2-4)^4} =[/m]
[m]=\frac{(8x^3-48x)(x^2-4) - (2x^4-24x^2) \cdot 4x}{(x^2-4)^3} =\frac{8x^5 - 48x^3 - 32x^3 + 192x - 8x^5 + 96x^3}{(x^2-4)^3} =\frac{16x^3 + 192x}{(x^2-4)^3}[/m]
6) [m]y=e^{-\frac{(x-1)^2}{2}}+1[/m]
[m]y'=e^{-\frac{(x-1)^2}{2}} \cdot (-\frac{(x-1)^2}{2})' +1'= -e^{-\frac{(x-1)^2}{2}} \cdot (x-1) = (1-x)e^{-\frac{(x-1)^2}{2}}[/m]
[m]y'' = (1-x)'e^{-\frac{(x-1)^2}{2}} + (1-x)(e^{-\frac{(x-1)^2}{2}})' = -e^{-\frac{(x-1)^2}{2}} + (1-x)(1-x)e^{-\frac{(x-1)^2}{2}} =[/m]
[m]= e^{-\frac{(x-1)^2}{2}}((1-x)^2 - 1) = e^{-\frac{(x-1)^2}{2}}(x^2-2x)[/m]