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

Question

Mathematics

Rex Hooper

23 July, 17:04

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

+2

Answers (2)

Know the Answer?

Shawn Cole · Answer 1

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

Fresier · Answer 2

The answer to this question is:

B) (sec x) ^2 + C

If dy/dx = tan (x), then y=A) 1/2 (tan x) ^2 + CB) (sec x) ^2 + CC) ln|sec x| + CD) ln|cos x| + CE) sec x tan x + C

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