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
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.
2 answers
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)
∂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)
1 answer