Implicit Differentiation: Finding the Derivative of sin(xy)

[kevinnn]

Homework Statement

I'm working on implicit differentiation and there is one part of the problem I'm having trouble with so I just pulled it out.
d/dx[6+sin(xy)]

Homework Equations

The Attempt at a Solution

They get the answer of this to be sin(xy) [(x) dy/dx+y] How do they get that? Here is what I get and how I arrive at it.
d/dx[6+sin(xy)]= d/dx[sin(xy)]= cos(xy) [d/dy (xy) ][dy/dx]= xcos(xy)dy/dx I believe the x comes out because we are evaluating the derivative of y at x, so the y is one but the x comes out, is that the right idea? Thanks.

-Kevin

 

[Fredrik]

They get the answer of this to be sin(xy) [(x) dy/dx+y]

Are you sure its sin, and not cos?

d/dx[6+sin(xy)]= d/dx[sin(xy)]= cos(xy) [d/dy (xy) ][dy/dx]

Where did the d/dy come from? I guess you must have rewritten d/dx as dy/dx d/dy, using the chain rule. That seems unnecessary, since it's not easier to apply d/dy to xy than to apply d/dx to it.

 

[kevinnn]

Yes sorry it is cos(xy)
Yes, I was using the chain rule. So if apply d/dx to it then the answer for that part would be y correct?

 

[SammyS]

Homework Statement

I'm working on implicit differentiation and there is one part of the problem I'm having trouble with so I just pulled it out.
d/dx[6+sin(xy)]

Homework Equations

The Attempt at a Solution

They get the answer of this to be [STRIKE]sin[/STRIKE] cos(xy) [(x) dy/dx+y] How do they get that? Here is what I get and how I arrive at it.
d/dx[6+sin(xy)]= d/dx[sin(xy)]= cos(xy) [d/dy (xy) ][dy/dx]= xcos(xy)dy/dx I believe the x comes out because we are evaluating the derivative of y at x, so the y is one but the x comes out, is that the right idea? Thanks.

-Kevin

You should have

$\displaystyle \frac{d}{dx}(\sin(xy))=\cos(xy)\frac{d}{dx}(xy)\ .$

Now, what is $\displaystyle \ \ \frac{d}{dx}(xy)\ ?$

 

[kevinnn]

Upon working through the problem again from the start I was able to solve it. I guess I was just tired when I first attempted it. Thanks for the help!