ese maestro nos pone a hacer esto ahora; Dice que son
Derivada de funciones trigonometricas directas
Expliquenme que no entiendo eso del coseno y seno como se resuelven:
a) (tanx)³
b) 2sen(3x+4)
c) y= cos² 1/1-x
d) y= senx/cosx
e) y= SenXCosX
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Answers & Comments
Verified answer
a)
y = (tan x)^3
y' = (sec x)^2 * 3 (tan x)^2
y' = 3 sec^2 x tan^2 x
b)
y = 2 sen (3x+4)
y' = 2 * 3 * cos (3x+4)
y' = 6 cos (3x+4)
c)
y = cos^2 (1/(1-x))
y' = [(0 - (-1))/(1-x)^2] * [-sen (1/(1-x))] * 2 cos (1/(1-x))
y' = [1/(1-x)^2] * [-sen (1/(1-x))] * 2 cos (1/(1-x))
y' = -2 sen (1/(1-x)) cos (1/(1-x)) / (1-x)^2
d)
y = sen x / cos x = tan x
y' = sec^2 x
e)
y = sen x cos x
y' = cos x cos x + sen x (-sen x)
y' = cos^2 x - sen^2 x
Suerte
a) d(tanx)³= 3(tan²x)sec²x
b) d(2sen(3x+4)= 2*3*sen(3x+4)=6sen(3x+4)
d) sen/cosx= tanx
d(tanx)= secx² =1+tan²x
e) d(senx cosx)=senx * (-senx)+cosx*cosx =-sen²x +cos²x