обратной тригонометрической функции 2
f(x)=sin3x
f(u)=sinu; u=3x
(sinu)`=(cosu)*(u)`
f`(x)=(sin3x)`=(cos3x)*(3x)`=(cos3x)*(3)=[b]3cos3x[/b]
2)
f(x)=cos(1-2x)
f(u)=cosu; u=(1-2x)
(cosu)`=(-sinu)*(u)`
f`(x)=(cos(1-2x))`=(-sin(1-2x))*(1-2x)`=(-sin(1-2x))*(-2)[b]=2sin(1-2x)[/b]
3)
[m]f`(x)=\frac{1}{cos^25x}\cdot (5x)`=\frac{5}{cos^25x}[/m]
4)
[m]f`(x)=-\frac{1}{sin^2(x-2)}\cdot (x-2)`=-\frac{1}{sin^2(x-2)}[/m]
5)
f`(x)=(sin(3-2x))`=(cos(3-2x))*(3-2x)`=(cos(3-2x))*(-2)=[b]-2cos(3-2x)[/b]
6)
[m]f`(x)=-\frac{1}{sin^2(5-3x)}\cdot (5-3x)`=-\frac{1}{sin^2(5-3x)}\cdot (-3)=\frac{3}{sin^2(5-3x)}[/m]