Posted by **James** on Tuesday, October 16, 2012 at 8:34pm.

Find the d^3z/dydx^2 partial derivative of:

ln[sin(x-y)]

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(x-y)/sin(x-y)

I got the d^2z/dydx derivative to be:

-(sin(x-y))^2 - (cos(x-y))^2/ (sin(x-y))^4

- Calculus -
**Steve**, Wednesday, October 17, 2012 at 10:29am
I think there's a minus sign missing there.

∂z/∂y = -cos(x-y)/sin(x-y) = -cot(x-y)

∂^2z/∂y∂x = csc^2(x-y)

∂^3z/∂y∂x^2 = 2csc(x-y) (-csc(x-y)cot(x-y))

= -2csc^2(x-y)cot(x-y)

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