f(x)=x2−9x+1−−−−−−√Решение. Способ 1f′(x)=(x2−9x+1−−−−−−√)′==(x2−9x+1)′2⋅x2−9x+1−−−−√==(x2−9)′⋅(x+1)−(x2−9)⋅(x+1)′(x+1)22⋅x2−9x+1−−−−√==(x2)′⋅(x+1)−(x2−9)⋅1(x+1)22⋅x2−9x+1−−−−√==2⋅x⋅(x+1)−(x2−9)(x+1)22⋅x2−9x+1−−−−√Ответ:
2⋅x⋅(x+1)−(x2−9)(x+1)22⋅x2−9x+√x^2-9)/(x+1