please help by using the composite rule .differentiate the function k(x)=sin(1/3x^3/2) (x>0)?

Answer 1

if k(x)=sin(1/3x^3/2)

then

k'(x) = (-1/2)cos(1/3x^3/2)x^(-5/2)

Answer 2

initiate with the 4. 4 equals 2 squared to boot as 2+2. then you take the three/2. 3 halves is what ex-different halves will take in case you get married and seperated thrice. So 3/2 & the 4 provides you sixteen. sturdy success.

Answer 3

if k(x)=sin(1/3x^3/2) (x>0),

just write down;
k'(x)
=(1/2)*sqrt(x)*cos(x^(3/2)/3)

i hope that this helps

Answer 4

Let y = sin u where u = (1/3).x^(3/2)
du / dx = (1/2).x^(1/2)
dy / du = cos u
dy / dx = (dy/du).(du/dx)
dy / dx = cos[(1/3).x^(3/2)]. (1/2).x^(1/2) = k ` (x)
(final expression cannot really be simplified much further apart from taking (1/2) to the front.)

Answer 5

k(x) = sin[(1/3)x^(3/2)](x>0)

k'(x)
= cos[(1/3)x^3/2][(1/3)(3/2)√x], chain rule
= cos[(1/3)x^3/2][(1/2)√x], simplified

Answer 6

(3/2 * 1/3x ^ (3/2-1)) * cos (1/3x^3/2)

I think

Answer 7

Go to "Start"......then click on "All programs"..... find "accessories" then click on "calculator".....then Wolla!