Posted by James on Tuesday, October 16, 2012 at 8:34pm.
Find the d^3z/dydx^2 partial derivative of:
ln[sin(xy)]
The answer is supposed to be 1/2; however, I cannot figure out how to get this answer. Any help is appreciated.

Calculus  James, Tuesday, October 16, 2012 at 8:36pm
I got the dz/dy derivative to be:
cos(xy)/sin(xy)
I got the d^2z/dydx derivative to be:
(sin(xy))^2  (cos(xy))^2/ (sin(xy))^4 
Calculus  Steve, Wednesday, October 17, 2012 at 10:29am
I think there's a minus sign missing there.
∂z/∂y = cos(xy)/sin(xy) = cot(xy)
∂^2z/∂y∂x = csc^2(xy)
∂^3z/∂y∂x^2 = 2csc(xy) (csc(xy)cot(xy))
= 2csc^2(xy)cot(xy)