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# Find $dy/dx$ by implicit differentiation.$\sin (xy) = \cos (x+y)$

## $$y^{\prime}=-\frac{y \cos (x y)+\sin (x+y)}{x \cos (x y)+\sin (x+y)}$$

Derivatives

Differentiation

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So for this problem, what we're gonna want to do is implicit differentiation and implicit differentiation is very good when we have some expression with, you know, an X and A Y, and it's equal to another expression. So we have this equation. Um, that's not just a simple asses. Why equals something? It's some function of X and Y perhaps and then another function of X and Y on the other side. So that's why we want to use implicit differentiation. So what we're given here is thesis ein of X y being equal to the co sign of X plus y. We're gonna differentiate both sides with respect tax. So it's gonna look something like this. So now when we do this, what we end up getting is we want to do the chain role. So, with the chain role we now have that this is the co sign of X y times. Why plus X y prime and that's doing chain role. That's gonna be equal to a negative sign of X plus y times one plus why prime? So with all of this, we want to simplify further. So we have now is why cosine X y plus x co sign X y times y prime And that's gonna be equal to the negative sign of X plus y minus the sign of X y x plus y white prime. So now that we have all this, we want toe ad, um, this right here to both sides and then we want to factor out the why prime. So when we do that, we get, um, X curse Sane X y plus sign X plus y We moved this over here and we moved this over here. So now that we have that, this is going to be all times. Why prime? And it's equal to a negative sign. X plus y minus. Why co sign X y and then we just divide the whole thing by this portion right here. So what we end up getting is that why prime is equal to a negative sign? Plus why, minus why could sign ex wife all over x co sign X y plus sign X plus y. This will be our final answer for the problem

California Baptist University

#### Topics

Derivatives

Differentiation

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp