13 uždavinys
f(x) =$$\sqrt {2}\cdot x^{2}+\sqrt {2}$$. Apskaičiuokite f'(√2)
Sprendimas.
( saknis(
2)* x^2+saknis(2))′ =
2)* x^2+saknis(2))′ = |
|
( saknis(2)* x^2+saknis(2))′ = $$(\sqrt {2}\cdot x^{2}+\sqrt {2})'$$ =
2)* x^2+saknis(2))′ = $$(\sqrt {2}\cdot x^{2}+\sqrt {2})'$$ = $${\normalsize (\sqrt {2}\cdot x^{2}+\sqrt {2})'}$$ = $${\normalsize (\sqrt {2}\cdot x^{2})'+\sqrt {2}'}$$
Paaiškinimas:
Paaiškinimas: Sumos išvestinė (f+g)′ = f′ + g′
( saknis(2)* x^2)′+ saknis(2)′ = $$(\sqrt {2}\cdot x^{2})'+\sqrt {2}'$$ =
2)* x^2)′+ saknis(2)′ = $$(\sqrt {2}\cdot x^{2})'+\sqrt {2}'$$ = $${\normalsize \sqrt {2}'}$$ = $$0$$ Paaiškinimas:
Paaiškinimas: Konstantos išvestinė yra 0
( saknis(2)* x^2)′+0 = $$(\sqrt {2}\cdot x^{2})'+0$$ =
2)* x^2)′+0 = $$(\sqrt {2}\cdot x^{2})'+0$$ = ( saknis(2)* x^2)′ = $$(\sqrt {2}\cdot x^{2})'$$ =
2)* x^2)′ = $$(\sqrt {2}\cdot x^{2})'$$ = $${\normalsize \sqrt {2}\cdot x^{2}}$$ = $${\normalsize \sqrt {2}\cdot (x^{2})'}$$ Paaiškinimas:
Paaiškinimas: $${\normalsize \sqrt {2}}$$ iškeltas prieš skliaustus
saknis(2)* ( x^2)′ = $$\sqrt {2}\cdot (x^{2})'$$ =
2)* ( x^2)′ = $$\sqrt {2}\cdot (x^{2})'$$ = $${\normalsize (x^{2})'}$$ = $${\normalsize 2\cdot x}$$ Paaiškinimas:
Paaiškinimas: Laipsnio išvestinė, kur n = 2
saknis(2)* 2* x = $$\sqrt {2}\cdot 2\cdot x$$ =
2)* 2* x = $$\sqrt {2}\cdot 2\cdot x$$ = Paaiškinimas: Keitimas $${\normalsize x}$$ = $${\normalsize \sqrt {2}}$$.
saknis(2)* 2* saknis(2) = $$\sqrt {2}\cdot 2\cdot \sqrt {2}$$ =
2)* 2* saknis(2) = $$\sqrt {2}\cdot 2\cdot \sqrt {2}$$ = saknis(2)* saknis(2)* 2 = $$\sqrt {2}\cdot \sqrt {2}\cdot 2$$ =
2)* saknis(2)* 2 = $$\sqrt {2}\cdot \sqrt {2}\cdot 2$$ = $${\normalsize \sqrt {2}\cdot \sqrt {2}}$$ = $${\normalsize 2}$$
2* 2 = $$2\cdot 2$$ =
$${\normalsize 2\cdot 2}$$ = $${\normalsize 4}$$
4$$4$$
$$(\sqrt {2}\cdot x^{2}+\sqrt {2})'$$ = $$$$
$$(\sqrt {2}\cdot x^{2})'+\sqrt {2}'$$ = $$$$
$$(\sqrt {2}\cdot x^{2})'$$ = $$$$
$$\sqrt {2}\cdot 2\cdot x$$ = $$$$
$$\sqrt {2}\cdot 2\cdot \sqrt {2}$$ = $$$$
$$4$$ $$$$
Atsakymas: 4