Текст задатака објашњених у видео лекцији:
Одредити извод функције:
Пр.7) y=√x+1x−1
Пр.8) y=ln3√2x−1
Пр.9) y=(x+(x+(x2+1)2)2)2
Пр.7) y=√x+1x−1
y′=(√x+1x−1)′=((x+1x−1)12)′=12(x+1x−1)12−1(x+1x−1)′=
=12(x+1x−1)−12(x+1)′(x−1)−(x+1)(x−1)′(x−1)2=12(x+1x−1)−12⋅(x−1)−(x+1)(x−1)2=
=12(x+1x−1)−12⋅(x−1)−(x+1)(x−1)2=12(x+1x−1)−12⋅2(x−1)2=√x+1x−1⋅1(x−1)2
Пр.8) y=ln3√2x−1
y′=13√2x−1⋅((2x−1)13)′=13√2x−1⋅13⋅(2x−1)13−1⋅(2x−1)′=
=13√2x−1⋅13⋅(2x−1)−23⋅2=23⋅1(2x−1)13(2x−1)23=23(2x−1)
Пр.9) y=(x+(x+(x2+1)2)2)2
y′=((x+(x+(x2+1)2)2)2)′=
=2⋅(x+(x+(x2+1)2)2)⋅(x+(x+(x2+1)2)2)′=
=2⋅(x+(x+(x2+1)2)2)⋅(1+((x+(x2+1)2)2)′)=
=2⋅(x+(x+(x2+1)2)2)⋅(1+2(x+(x2+1)2)⋅(x+(x2+1)2)′)=
=2⋅(x+(x+(x2+1)2)2)⋅(1+2(x+(x2+1)2)⋅(1+((x2+1)2)′))=
=2⋅(x+(x+(x2+1)2)2)⋅(1+2(x+(x2+1)2)(1+2(x2+1)(x2+1)′))=
=2⋅(x+(x+(x2+1)2)2)⋅(1+2(x+(x2+1)2)(1+2(x2+1)2x))