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 $dy/dx=-f_x/f_y$
$dy/dx=- (2sinxcosx-16)/(2sinycosy+16)$

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