Implicit Diff Problem. Whao made this stuff up

**psilver1** · October 4th 2012, 06:51 PM

I have tried so many ways and I am sure I am over analyzing these problems but I dont know if I am suppose to use the Chain Rule Product Rule, I am becoming more confused and just need a step by step visual to get me jump started. If you have time could you please explain why each thing takes place. If not I understand, anyway here is the question.

Find DxY by Implicit Diff

xy + sin(xy) = 1

**TheEmptySet** · October 4th 2012, 07:19 PM

Originally Posted by psilver1

I have tried so many ways and I am sure I am over analyzing these problems but I dont know if I am suppose to use the Chain Rule Product Rule, I am becoming more confused and just need a step by step visual to get me jump started. If you have time could you please explain why each thing takes place. If not I understand, anyway here is the question.

Find DxY by Implicit Diff

xy + sin(xy) = 1

$\frac{d}{dx}(xy)+\frac{d}{dx}\sin(xy)=\frac{d}{dx} 1$

$y \frac{d}{dx}x+x\frac{d}{dx}y+\cos(xy)\frac{d}{dx}( xy)=0$

$y \frac{d}{dx}x+x\frac{d}{dx}y+\cos(xy)(y\frac{d}{dx }x+x\frac{d}{dx}y)=0$

$y+x\frac{dy}{dx}+\cos(xy)(y+x\frac{dy}{dx})=0$

Now just solve for the derivative.

**MarkFL** · October 4th 2012, 07:30 PM

For the first term, we may use the product rule, for the second term the chain rule, then the product rule, and the right side is of course zero:

$x\cdot\frac{dy}{dx}+y+\cos(xy)\left(x\cdot\frac{dy }{dx}+y \right)=0$

$\left(x\cdot\frac{dy}{dx}+y \right)(1+\cos(xy))=0$

What does this imply?

**psilver1** · October 4th 2012, 08:18 PM

So (xy' + y)(xy'cos(xy) + ycos(xy))=0
sorta stuck does the xy' factor out. (xy' + y)(cos(xy) + ycos(xy))=0

thanks for the help so far. the above is where I am screwing this thing up. Mark I dont see where the +cos(xy)[x * dy/dx + y] turns into 1 + cos(xy). Btw I am sure you are right, i just dont understand where it comes from is what I am trying to say

**richard1234** · October 4th 2012, 08:26 PM

Originally Posted by psilver1

So (xy' + y)(xy'cos(xy) + ycos(xy))=0
sorta stuck does the xy' factor out. (xy' + y)(cos(xy) + ycos(xy))=0

thanks for the help so far. the above is where I am screwing this thing up. Mark I dont see where the +cos(xy)[x * dy/dx + y] turns into 1 + cos(xy). Btw I am sure you are right, i just dont understand where it comes from is what I am trying to say

Look closely, Mark simply factored the whole thing. It's correct.

**psilver1** · October 4th 2012, 09:09 PM

Ahh I see it. So is 1+cos(xy) and identity. Sorry for the questions just what to clear my mind.

**MaxJasper** · October 4th 2012, 09:20 PM

xy+sin(xy)-1 as surface:

**MarkFL** · October 4th 2012, 09:27 PM

It is simply a factor we may ignore, as it doesn't involve $\frac{dy}{dx}$ , and we know it cannot be zero, since this would imply:

$xy=(2k+1)\pi$

But, if this is true, then the original equation becomes:

$xy=1$

Which is a contradiction.

**psilver1** · October 4th 2012, 09:35 PM

So ignoreing the (1+ cos(xy). We are left with xy' + y = 0 which would give y'= - y/x. Which is the correct answer but where does the cosxy move away too.

**MarkFL** · October 4th 2012, 10:01 PM

It doesn't move away, we just ignore it as a factor that cannot be zero.

If you use the method suggested by TheEmptySet, you will find this factor is divided out or cancels. I highly suggest you try this method as well.

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