# Implicit Diff Problem. Whao made this stuff up

Printable View

• October 4th 2012, 07:51 PM
psilver1
Implicit Diff Problem. Whao made this stuff up
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
• October 4th 2012, 08:19 PM
TheEmptySet
Re: Implicit Diff Problem. Whao made this stuff up
Quote:

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.
• October 4th 2012, 08:30 PM
MarkFL
Re: Implicit Diff Problem. Whao made this stuff up
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?
• October 4th 2012, 09:18 PM
psilver1
Re: Implicit Diff Problem. Whao made this stuff up
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
• October 4th 2012, 09:26 PM
richard1234
Re: Implicit Diff Problem. Whao made this stuff up
Quote:

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.
• October 4th 2012, 10:09 PM
psilver1
Re: Implicit Diff Problem. Whao made this stuff up
Ahh I see it. So is 1+cos(xy) and identity. Sorry for the questions just what to clear my mind.
• October 4th 2012, 10:20 PM
MaxJasper
Re: Implicit Diff Problem. Whao made this stuff up
• October 4th 2012, 10:27 PM
MarkFL
Re: Implicit Diff Problem. Whao made this stuff up
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.
• October 4th 2012, 10:35 PM
psilver1
Re: Implicit Diff Problem. Whao made this stuff up
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.
• October 4th 2012, 11:01 PM
MarkFL
Re: Implicit Diff Problem. Whao made this stuff up
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.