My Puzzling Work on Differentiating a Fraction

[snowJT]

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 d(\frac{x^2}{y})

this was my work...

d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2
d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}

I'm just confused because I would of thought that it would of been something like this...

d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2

hmm...?

 

[HallsofIvy]

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 d(\frac{x^2}{y})

this was my work...

d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2
d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}

I'm just confused because I would of thought that it would of been something like this...

d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2

hmm...?

What do you mean by "IC"? When I saw "Solving an IC" in the differential equation forum, I thought you mean an "initial condition" problem but I see no differential equation at all here!

Certainly, the derivative of x²/y is, by the quotient rule,
\frac{2xy- x^2}{y^2}
There is no "y" in the second term in the numerator. (I surely don't see where you get your first line
d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2
that makes no sense to me.)

Of course, you could also write x²/y as x²y^-1 and use the product rule. In that case, the dervative is
2xy^{-1}+ (-1)x^2y^{-2}
Just rewrite that as
\frac{2x}{y}- \frac{x^2}{y^2}
multiply the numerator and denominator of the first fraction by y and combine and you get the same answer as with the quotient rule.

Since this has nothing to do with differential equations, I moving it to calculus.

 

[snowJT]

sorry, I didn't know the difference, this was on a DE test

but thanks for explaining it