Če f (x) = cot2 x in g (x) = e ^ (1 - 4x), kako ločiti f (g (x)) z verigo?

Odgovor:

# (8e ^ (1-4x)) / sin ^ 2 (2e ^ (1-4x)) # ali # 8e ^ (1-4x) csc ^ 2 (2e (1-4x)) #

Pojasnilo:

#f (g (x)) = cot2e ^ (1-4x) #

Let #g (x) = u #

#f '(u) = d / (du) cot2u = d / (du) (cos2u) / (sin2u) = (- 2sin (2u) sin (2u) -2cos (2u) cos (2u)) / sin ^ 2 (2u) #

# = (- 2sin ^ 2 (2u) -2cos ^ 2 (2u)) / sin ^ 2 (2u) #

# = - 2 / sin ^ 2 (2u) #

#g '(x) = - 4e ^ (1-4x) #

Z uporabo verižnega pravila: #f '(g (x)) = f' (u) * g '(x) #

# = - 2 / sin ^ 2 (2u) * - 4e ^ (1-4x) #

# = - 2 / sin ^ 2 (2e ^ (1-4x)) * - 4e ^ (1-4x) #

# = (8e ^ (1-4x)) / sin ^ 2 (2e ^ (1-4x)) # ali # 8e ^ (1-4x) csc ^ 2 (2e (1-4x)) #

Kako ločiti naslednjo parametrično enačbo: x (t) = t / (t-4), y (t) = 1 / (1-t ^ 2)?

Dy / dx = - (t (t-4) ^ 2) / (2 (1-t ^ 2) ^ 2) = - t / 2 ((t-4) / (1-t ^ 2)) ^ 2 dy / dx = (y '(t)) / (x' (t)) y (t) = 1 / (1-t ^ 2) y '(t) = ((1-t ^ 2) d / dt [1] -1d / dt [1-t ^ 2]) / (1-t ^ 2) ^ 2 barva (bela) (y '(t)) = (- (- 2t)) / (1-t ^ 2) ^ 2 barva (bela) (y '(t)) = (2t) / (1-t ^ 2) ^ 2 x (t) = t / (t-4) x' (t) = ((t -4) d / dt [t] -td / dt [t-4]) / (t-4) ^ 2 barva (bela) (x '(t)) = (t-4-t) / (t 4) ^ 2 barva (bela) (x '(t)) = - 4 / (t-4) ^ 2 dy / dx = (2t) / (1-t ^ 2) ^ 2 -: - 4 / (t -4) ^ 2 = (2t) / (1-t ^ 2) ^ 2xx- (t-4) ^ 2/4 = (- 2t (t-4) ^ 2) / (4 (1-t ^ 2) ) ^ 2) = - (t (t-4) ^

Če f (x) = cos5 x in g (x) = e ^ (3 + 4x), kako ločiti f (g (x)) z verigo?

Leibnizova notacija je lahko uporabna. f (x) = cos (5x) Naj bo g (x) = u. Potem izpeljava: (f (g (x))) '= (f (u))' = (df (u)) / dx = (df (u)) / (dx) (du) / (du) = (df (u)) / (du) (du) / (dx) = = (dcos (5u)) / (du) * (d (e ^ (3 + 4x))) / (dx) = = -sin (5u) * (d (5u)) / (du) * e ^ (3 + 4x) (d (3 + 4x)) / (dx) = = -sin (5u) * 5 * e ^ (3 + 4x) ) * 4 = = -20sin (5u) * e ^ (3 + 4x)

Če f (x) = cos 4 x in g (x) = 2 x, kako ločiti f (g (x)) z verigo?

-8sin (8x) Pravilo verige je navedeno kot: barva (modra) ((f (g (x))) '= f' (g (x)) * g '(x)) Najdemo derivat f ( x) in g (x) f (x) = cos (4x) f (x) = cos (u (x)) Uporabiti moramo verigo na f (x) vedeti, da (cos (u (x)) ' = u '(x) * (cos' (u (x)) Naj bo u (x) = 4x u '(x) = 4 f' (x) = u '(x) * cos' (u (x)) barva (modra) (f '(x) = 4 * (- sin (4x)) g (x) = 2x barva (modra) (g' (x) = 2) Nadomestitev vrednosti na lastnost: barva (modra) ) ((f (g (x))) '= f' (g (x)) * g '(x)) (f (g (x)))' = 4 (-sin (4 * (g (x) ))) * 2 (f (g (x))) '= 4 (-sin (4 * 2x)) * 2 (f (g (