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)
Answer This Question
Related Questions
 Calculus  My question is:Use L'Hospital's rule to find limit as x approaches 0...
 Calculus 12th grade (double check my work please)  1.)Find dy/dx when y= Ln (...
 Calculus  If x = t^2 + 1 and y = t^3, then d^2y/dx^2 = I know I can solve for t...
 Math  derivative of sinusoidal (check)  a) Find a function y=f(x) that ...
 Calculus II  y = *The Integral Of sin(4e^t)dt* a = 0 b = ln(2x) I'm supposed to...
 Is this f'(x) or partial derivative?  Find d VC/ d y = ((y/240)^2)(w) = 2yw/240...
 Calculus  Hello, Could somebody kindly check my answer for the following ...
 Calculus  Use implicit differentiation to find dy/dx. e^4x = sin(x+2y). This is...
 calculus  Please help. Applying the chain rule, how do I find the derivative of...
 Calculus  Any help would be much appreciated with the steps involved in each ...
More Related Questions