Partial derivative

answered 2014-07-17 13:28:30 -0600

Justin
1009 ●2 ●20 ●38

Hm, I'm not completely sure about how you took your derivatives, but let me show you my steps for using the product rule on this derivative:

$$ f(x, y) = (x^2 - y) * (y^2-x)^{-1} $$

$$ \frac{\partial f}{\partial x} = (2x) * (y^2-x)^{-1} + (-1)(y^2 - x)^{-2}(-1)*(x^2 - y) $$ $$ = \frac{2x}{y^2-x} + \frac{x^2 - y}{(y^2 - x)^2}$$

Multiplying the first term by $\frac{y^2-x}{y^2-x}$ yields:

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

WolframAlpha is telling me that it got my answer as well (I'm not sure what it's telling you, although it looks to be the negative version of what I got). More importantly though, I had to do much more work with the product rule than with the quotient rule here.

For reference, my calculus teacher's mnemonics for the product rule and quotient rule are:

Product Rule: One prime two plus two prime one, isn't calculus just such fun?

Quotient Rule: Lo dee hi less hi dee lo, denominator squared we go! (denominator times derivative of numerator minus numerator times derivative of denominator with the denominator squared)

answered 2014-07-17 16:01:54 -0600

Dylan
650 ●2 ●16 ●25

Spiff, I believe your method of derivative is equivalent to the quotient rule. The quotient rule states that for two functions $f$ and $g$ (for one variable), that

$$\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{f}{g} \right) = \frac{f'g-g'f}{g^2}$$

If we write this in Spiff's manner, we get

$$\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{f}{g} \right) = \frac{\mathrm{d}}{\mathrm{d}x} \left(f g^{-1} \right),$$

which according to the product rule is just this:

$$f'g^{-1}+(g^{-1})'f$$

If we algebraically rewrite this a bit, and use the chain rule on $(g^{-1})'$, then we can express this as follows:

$$\frac{f'}{g} + \left(\frac{-1}{g^2}\right) g'f$$

$$=\frac{gf'}{g^2}-\frac{g'f}{g^2}$$

$$= \frac{f'g -g'f}{g^2}.$$

So it seems like you can derive the quotient rule from your method. I actually like this a lot, and assuming I didn't make any mistakes, I will probably use this myself.

As far as the problem goes, Wolfram Alpha usually does some "simplification" that I don't always find easy to replicate, so I usually just try plugging in what I got and see if an alternate form matches their answer.