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y = sin(x+y)

cos( x + y )[(1 + y')] = y'

cos(x + y ) + y'cos(x + y ) = y'

y'-y'cos( x+ y) = cos( x + y )

y'[1-cos(x+y)]= cos(x+y)

y'= [cos(x+y)]/ [1-cos(x+y)]

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βˆ™ 12y ago
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Q: What is the implicit differentiation of y equals sin x plus y?
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