Задача 31353 #8. 25...

sanchoys2310

2018-11-27 20:01:59

#8. 25

sova

2018-11-27 22:13:05

(u/v)`=(u`*v-u*v`)/v^2

u=cos^2x
u`=2cosx*(cosx)`=2cosx*(-sinx)=-sin2x

v=lg(x^2-2x+1)
v`=(1/(x^2-2x+1)*ln10)*(x^2-2x+1)`=(2x-2)/((x^2-2x+1)*ln10)=

=2/((x-1)*ln10)

О т в е т. (-sin2x*lg(x^2-2x+1)- 2cos^2x/((x-1)*ln10))/lg^2(x^2-2x+1)