How do you differentiate #(sin^2x+sin^2y)/(x-y)=16#?

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Answer 1 · 2016-11-03T19:13:54

#dy/dx=- (2sinxcosx-16)/(2sinycosy+16)#

#(sin^2x+sin^2y)/(x-y)=16#

#sin^2x+sin^2y=16(x-y)#

#sin^2x+sin^2y=16x-16y#

#sin^2x+sin^2y-16x+16y=0#

Take the derivative with respect to x and hold y constant.
#f_x=2sinxcosx-16#

Take the derivative with respect to y and hold x constant
#f_y=2sinycosy+16#

To write your final answer use #dy/dx=-f_x/f_y#
#dy/dx=- (2sinxcosx-16)/(2sinycosy+16)#