Let f(x) = cosx. Then f'''(pi/6) equals...?

Question

Either:
1/2
-1/2
-(3^1/2)/2
(3^1/2)/2

Does ''' meaning do the derigative of it three times, or is that something else all together? Could somebody please go through this step by step on how to find the answer? I'm just as concerned with learning how to actually go about this as I am of getting it right. Thank you.

Ah, I get it. That seems so easy now that you have explained it. Thanks.

Answer 1

yes, f'''(x) means the third derivative.

f(x) = cos(x)
f'(x) = -sin(x)
f''(x) = -cos(x)
f'''(x) = sin(x)

so f'''(pi/6) = sin(pi/6) = 1/2

Jimmy got the derivatives wrong. (sin x)' = cos x, (cos x)' = -sin x.

Answer 2

Find 1st, 2nd and 3rd derivative

f(x)= cosx
f'(x)= (-sinx)
f''(x)= (-cosx)
f'''(x)= sinx
f'''(pi/6) = sin (pi/6)= 1/2
f'''(1/2) = sin (1/2)= 0.48
f'''(-1/2) = sin (-1/2)= 0.48
f'''-(3^1/2)/2 = sin -(3^1/2)/2= - 0.68
f'''(3^1/2)/2 = sin (3^1/2)/2= 0.68
______________
1/2
-1/2
-(3^1/2)/2
(3^1/2)/2

Change to radian to calculate!

Answer 3

cos(x)>sin(x)>-cos(x) pi over 6 is the unit cirlce indicated Srt(3/2) , (1/2)... it is either negitve half or sqrt root 3/2 I forgot... I think negitive half.

Answer 4

(sin(x))' = -cos(x)
(cos(x))' = sin(x)

so,
f(x) = cos(x)
f'(x) = sin(x)
f''(x) = -cos(x)
f'''(x) = -sin(x)
f'''(pi/6) = -sin(pi/6) = -1/2