second derivative help

[wvcogs]

Hi, any help with this would be great!

Find y" if sin(xy) = x^2 - y

I can get to here cos(xy)(xy'+y) = 2x - y'
using implicit differentiation. I am having trouble collecting my y' terms. Plus I then need to do it again for y".

 

[Daniel_Feldman]

Hello,

The problem is

sin(xy) = x^2 - y

Your implicit differentiation

cos(xy)(xy'+y) = 2x - y'

is correct.

Let's see where we can go from here.

Multiplying out the left side yields

xy'cos(xy)+ycos(xy)=2x-y'

Now, let's move ycos(xy) to the right side and y' to the left side, so that we can have all terms with y' one one side, and everything else on the other.

xy'cos(xy)+y'=2x-ycos(xy)

Now, we can factor out y' on the right side, which yields

(xcos(xy)+1)y'=2x-ycos(xy)

Now, we can divide:

y'=(2x-ycos(xy))/(xcos(xy)+1)

there's the first derivative, simplified.

y'' might get a bit messy here, but let's try it.....

Using the quotient rule for the right side, we get:

y''=[(xcos(xy)+1)(2-(-ysin(xy)(xy'+y)+y'cos(xy))-(2x-ycos(xy))(-xsin(xy)(xy'+y)+cos(xy))]/(xcos(xy)+1)^2

Yep, pretty messy. There's got to be an easier way. Let me think for a bit, and see if you can work with this in the meantime.

 

[wvcogs]

Thanks! I had worked on it some more and got that far with the y'. The y" seemed like it was going to be a mess, so I might wait until morning to look at it some more.
If I use the q rule, it is going to be huge. I wonder if there is something I am missing. Perhaps a way to pull something out as a factor?

Well, thanks.