foster4140203
foster4140203 foster4140203
  • 26-04-2017
  • Mathematics
contestada

If dy/dx = tan(x), then y=
A) 1/2 (tan x)^2 + C
B) (sec x)^2 + C
C) ln|sec x| + C
D) ln|cos x| + C
E) sec x tan x + C

Respuesta :

MissPhiladelphia
MissPhiladelphia MissPhiladelphia
  • 10-05-2017
dy/dx= tanx, can be answered directly using the derivatives of trigonometric functions but this is how the answer is derived
         =(sinx/cosx) basic trigonometric function
         = [cosx cox+sinxsinx]/cos^2x
         =[cos^2x+sin^2x]/cos^2x
cos^2+sin^2x = 1 ; fundamental trigonometric identities
         = 1/cos^2x; reciprocal relations
          = sec^2x+C
The answer is letter B.sec^2x+C
Answer Link
Аноним Аноним
  • 11-05-2017
The answer to this question is:

B) (sec x)^2 + C
Answer Link

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