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

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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).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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