Задача 56330 Помогите пожалуйста #2,#3...
Условие
Решение
[m]y`=(ln(2x-1))`=\frac{1}{2x-1}\cdot (2x-1)`=\frac{2}{2x-1}[/m]
[m]y``=(\frac{2}{2x-1})`=2\cdot ((2x-1)^{-1})`=2\cdot (-1)\cdot (2x-1)^{-1-1}\cdot (2x-1)`=-\frac{2}{(2x-1)^2}\cdot (2)=-\frac{4}{(2x-1)^2}[/m]
[m]y```=(-\frac{4}{(2x-1)^2})`=(-4\cdot(2x-1)^2)`=-4\cdot (-2)\cdot (2x-1)^{-3}\cdot (2x-1)`=\frac{16}{(2x-1)^3}[/m]
[m]y```=(\frac{16}{(2x-1)^3})`=16\cdot (-3)\frac{1}{(2x-1)^4}\cdot (2x-1)`=-\frac{16}{(2x-1)^4}[/m]
[m]y^{(5)}=-16\cdot (-4)\cdot \frac{1}{(2x-1)^5}\cdot (2x-1)=\frac{128}{(2x-1)^5}[/m]
3.
z`_(x)=(x^3+3x^2y-y^3)`_(x)=3x^2+6xy
z`_(y)=(x^3+3x^2y-y^3)`_(y)=3x^2-3y^2
dz=z`_(x)dx+z`_(y)dy
[b]dz=(3x^2+6xy)dx+(3x^2-3y^2)dy
[/b]
z``_(xx)=(3x^2+6xy)`_(x)=6x+6y
z``(xy)=(3x^2+6xy)`_(y)=6x
z``_(yy)=(3x^2-3y^2)`_(y)=-6y
d^2z=z``_(xx)dx^2+2z``_(xy)dxdy+z``_(yy)dy^2
[b]d^2z=(6x+6y)dx^2+2*6xdxdy-6ydy^2[/b]