Easy implicit differentiation (1 Viewer)

[*fkr]

Gday, we’ve only done bits and pieces of graphs but I think we’re supposed to know implicit differentiation. Think I was away. Could someone please tell me where the last x dy/dx term comes from in:<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>
<o> </o>

x<SUP>2</SUP> + y<SUP>2 </SUP>= xy + 3<o></o>​

2x + 2y dy/dx = y + x dy/dx<o></o>​

<o> </o>​

It’s an example from <?xml:namespace prefix = st1 ns = "urn:schemas-microsoft-comffice:smarttags" /><st1lace w:st="on"><st1:City w:st="on">Arnold</st1:City></st1lace>, thanks in advanced<o></o>

 

[acmilan]

Because x and y are both in terms of x, xy is basically a product of two things that can be written in terms of x. So you use product rule, (xy)' = x'y + xy' = y + xdy/dx

 

[pLuvia]

x² + y² = xy + 3
2x+2y*dy/dx = y + x*dy/dx
2y*dy/dx - x*dy/dx = y - 2x
dy/dx = (y-2x)/(2y-x)

 

[Riviet]

x² + y² = xy + 3
2x+2y*dy/dx = y + x*dy/dx
2x*dy/dx - x*dy/dx = y - 2x
dy/dx = (y-2x)/(2y-x)

Just a minor correction, the x in red should be a y.

 

[pLuvia]

Just a minor correction, the x in red should be a y.

Whoops, yeh I went back editting another part and must of forgot that