Задача 55405 Производны...
Условие
Решение
[m]y`=\frac{1}{x+y}\cdot (x+y)`[/m]
[m]y`=\frac{1}{x+y}\cdot (x`+y`)[/m]
x`=1
[m]y`=\frac{1}{x+y}+\frac{y`}{x+y}[/m]
[m]y`-\frac{y`}{x+y}=\frac{1}{x+y}[/m]
[m]y`(1-\frac{1}{x+y})=\frac{1}{x+y}[/m]
[m]y`\cdot \frac{x+y-1}{x+y}=\frac{1}{x+y}[/m]
[m]y`=\frac{1}{x+y}:\frac{x+y-1}{x+y}[/m]
[m]y`=\frac{1}{x+y-1}[/m]
[m]y``=(\frac{1}{x+y-1})`[/m]
[m]y``=-\frac{1}{(x+y-1)^2}\cdot (x+y-1)`[/m]
[m]y``=-\frac{1}{(x+y-1)^2}\cdot (1+y`)[/m]
[m]y``=-\frac{1}{(x+y-1)^2}-\frac{y`}{(x+y-1)^2}[/m]
[m]y``=-\frac{1}{(x+y-1)^2}-\frac{\frac{1}{x+y-1}}{(x+y-1)^2}[/m]
[m]y``=-\frac{1}{(x+y-1)^2}-\frac{1}{(x+y-1)^3}[/m]
[m]y``=-\frac{x+y-1}{(x+y-1)^3}-\frac{1}{(x+y-1)^3}[/m]
[m]y``=-\frac{x+y}{(x+y-1)^3}[/m]
2.
[m](x^3-y^2)`=(x^3)`\cdot y^3+x^3\cdot (y^3)`[/m]
[m]3x^2-2y\cdot y`=3x^2\cdot y^3+x^3\cdot 3y^2\cdot y`[/m]
[m]3x^2-3x^2\cdot y^3=2y\cdot y`+x^3\cdot 3y^2\cdot y`[/m]
[m]3x^2-3x^2\cdot y^3=(2y+x^3\cdot 3y^2)\cdot y`[/m]
[m]y`=\frac{3x^2-3x^2\cdot y^3}{2y+x^3\cdot 3y^2}[/m]