What is the derivative of #cos(x-y)=0#?

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Answer 1 · 2016-09-17T18:00:40

#dy/(dx)=1#

Given #f(x,y)=0#

#df=f_xdx+f_ydy=0# so

#dy/(dx)=-f_x/(f_y)= sin(x-y)/sin(x-y)=1#

Answer 2 · 2016-09-17T18:05:00

#dy/dx=1#

We can rearrange from the beginning:

#x-y=arccos(0)#

#x-y=pi/2#

#y=x-pi/2#

#dy/dx=1#