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

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Homework Help Overview

The discussion revolves around finding the derivative of the equation y = sin(x + y), exploring implicit differentiation and the context of partial derivatives.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to clarify the variable with respect to which the derivative is being taken. There are attempts to apply implicit differentiation and partial derivatives, with some participants questioning the original poster's intent.

Discussion Status

The discussion is active, with various approaches being explored, including implicit differentiation and partial derivatives. Some participants have provided guidance on how to approach the differentiation, while others are seeking clarification on the original question.

Contextual Notes

There is a noted ambiguity regarding the variable of differentiation, as well as a potential misunderstanding of the original poster's request.

mattsoto
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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 [itex]f(x,y) = sin(x+y)[/itex] are [itex]\frac{df}{dx} = cos(x+y)[/itex] and [itex]\frac{df}{dy} = cos(x+y)[/itex]
 
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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