Задача 57735 Математика...
Условие
Решение
y`=6*(2x^3+3x)^5*(2x^3+3x)`=6*(2x^3+3x)^5*(6x^2+3)
2.
y`=7*(x^3+x)^6*(x^3+x)`=7*(x^3+x)^6*(3x^2+1)
3.
y`=3sin^2x*(sinx)`=3sin^2x*cosx
4.
y`=(-sin(5x+2))*(5x+2)`=(-sin(5x+2))*(5)=-5sin(5x+2)
5.
y`=[m]\frac{1}{3+2x^2}[/m]*(3+2x^2)`=[m]\frac{1}{3+2x^2}[/m]*(0+4x)=[m]\frac{4x}{3+2x^2}[/m]
6.
y`=e^(3x-2)*(3x-2)`=e^(3x-2)*(3)=3*e^(3x-2)
7.
y`=5^(3x+2)*ln5*(3x+2)`=5^(3x+2)*ln5*(3)=3*(ln5)*5^(3x+2)
8.
y`=[m]\frac{1}{cos^2x^2}[/m]*(x^2)`=[m]\frac{1}{cos^2x^2}[/m]*(2x)=[m]\frac{2x}{cos^2x^2}[/m]
9.
y`=3sin^2(5x-7)*(sin(5x-7))`=3sin^2(5x-7)*cos(5x-7)*(5x-7)`=3sin^2(5x-7)*cos(5x-7)*(5)=15sin^2(5x-7)*cos(5x-7)
10.
y`=[m]\frac{1}{\sqrt{1-(2x+5)^2}}[/m]*(2x+5)`=[m]\frac{1}{\sqrt{1-(2x+5)^2}}[/m]*(2)=[m]\frac{2}{\sqrt{1-(2x+5)^2}}[/m]