What is the derivative of y=sin(x+y)?

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The derivative of the equation y = sin(x + y) can be calculated using implicit differentiation. The partial derivatives are given as ∂f/∂x = cos(x + y) and ∂f/∂y = cos(x + y). When differentiating y with respect to x, the result is y' = cos(x + y) / (1 - cos(x + y)). Conversely, when differentiating x with respect to y, the result is x' = (1 - y cos(x + y)) / cos(x + y). These results are crucial for understanding the relationship between x and y in this context.

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what is the derivative of y=sin(x+y)?

what is the derivative of y=sin(x+y)?
 
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The partial derivatives of f(x,y) = sin(x+y) are \frac{df}{dx} = cos(x+y) and \frac{df}{dy} = cos(x+y)
 
mattsoto - What do you want the derivative with respect to? If it's anything other than x or y, we need to talk a bit more.
 
the derivative is respect to y, Diane...
 
The derivative of y, given y= sin(x+ y), with respect to x, using implicit differentiation: y'= cos(x+y)(1+ y') so y'- y'cos(x+y)= cos(x+ y) and
y'= cos(x+y)/(1- cos(x+y)).

The derivative of x, given y= sin(x+ y), with respect to y (which is what you told you Diane you want, but I doubt since I would read what you originally wrote as 'the derivative OF y= ...), by implicit differentiation: 1= cos(x+y)(x'+ y) so
1- ycos(x+y)= cos(x+y)x' and x'= (1- ycos(x+y))/cos(x+y).
 

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