- #1
snowJT
- 117
- 0
Its funny.. because I got this right on a test.. but.. I'm looking back at it.. and it doesn't make sense how I figured it out...
I had to find the IC of [tex]d(\frac{x^2}{y})[/tex]
this was my work...
[tex]d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2[/tex]
[tex]d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}[/tex]
I'm just confused because I would of thought that it would of been something like this...
[tex]d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2[/tex]
hmm...?
I had to find the IC of [tex]d(\frac{x^2}{y})[/tex]
this was my work...
[tex]d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2[/tex]
[tex]d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}[/tex]
I'm just confused because I would of thought that it would of been something like this...
[tex]d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2[/tex]
hmm...?