How do you differentiate #xcosy+ycos x=1#?

1 Answer
Mar 10, 2015

Differentiate #xcosy+ycos x=1#

I assume that you want #(dy)/(dx)#, which is

#(dy)/(dx)=(ysinx-cosy)/(cosx-xsiny)#.

Method:
Since we can't (or don't want to) solve for #y#, leave the function(s) implicit and use implicit differentiation. (An application of the chain rule.)

Given: #xcosy+ycos x=1#
We know: #d/(dx)(xcosy+ycos x)= d/(dx)(1)#
So, using the product rule twice on the left:
#[(dx)/(dx)cosy+xd/(dx)(cosy)]+[(dy)/(dx)cos x+yd/(dx)(cosx)]=0#

Thus, we see that:

#cosy-xsiny (dy)/(dx)+(dy)/(dx)cosx-ysinx=0#

Solve algebraically for #(dy)/(dx)#, to get:

#(dy)/(dx)=(ysinx-cosy)/(cosx-xsiny)#.