# Finding $\frac{d}{dx} \frac{x^2}{y}$

$$\frac{d}{dx} \frac{x^2}{y}$$

According to Wolframalpha

I "factor out constants"

$$\frac{\frac{d}{dx} x^2}{y}$$

Then I will get $\frac{2x}{y}$. Is that right? But $y$ is not a constant? What I did actually (quotient rule got me stuck)

The actual question is "Find $\frac{d^2y}{dx^2}$ of $2x^3 - 3y^2 = 8$"

I got

$$\frac{dy}{dx} = \frac{x^2}{y}$$

Then

$$\frac{d^2y}{dx^2} = \frac{y \cdot 2x - x^2 \cdot \frac{dy}{dx}}{y^2}$$

$$= \frac{2xy - x^2 \cdot \frac{x^2}{y}}{y^2}$$

$$= \frac{2xy^2 - x^4}{y^3}$$

Is this correct? It doesn't look like a "simple" answer (or whats in wolfram)?

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@Deven The actual question is asking to find the $y''$ along a curve defined implicitly. So $y$ is not constant with respect to $x$. – alex.jordan Feb 26 '12 at 0:33
Ah, I didn't read "the actual question" part I just looked at the expression at the top of the page, my apology. – Deven Ware Feb 26 '12 at 0:38
+1 You $\TeX$ed in everything! – user21436 Feb 26 '12 at 0:47
Are you sure it didn't say "Find $\dfrac{d^2 y}{dx^2}$ if $2x^3-3y^2=8$"? – Michael Hardy Feb 26 '12 at 1:27

Your first attempt (second in order presented) is correct. There is some simplification that you can do though. Since you will only ever be evaluating this expression for $(x,y)$ on the originally defined curve, you can use that relation to simplify. Some "simplifications" would be

\begin{align*} \frac{d^2y}{dx^2} & = \frac{2xy^2-x^4}{y^3}&\frac{d^2y}{dx^2} & = \frac{2xy^2-x^4}{y^3}\\ & = \frac{2x(3y^2)-3x^4}{3y^2y}&& = \frac{x(4y^2-2x^3)}{2y^3}\\ & = \frac{2x(2x^3-8)-3x^4}{(2x^3-8)y}&& = \frac{x(4y^2-(8+3y^2))}{2y^3}\\ & = \frac{x^4-16x}{(2x^3-8)y}&& = \frac{x(y^2-8)}{2y^3}\\\\ & = \frac{x(x^3-16)}{2(x^3-4)y}\\ \end{align*}

This is only "simpler" in that lower powers of $y$ (respectively $x$) are used.

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 Right: it looks like what the OP entered into Wolfram confused it because no dependence of y on x was given; certainly that can be changed, for example, by saying "d/dx ((x^2)/y(x))" or some such. – Tyler Feb 26 '12 at 0:39 Is this really about elliptic curves? These different factored forms make it apparent where the inflection points are. – alex.jordan Feb 26 '12 at 0:49 Haha, ok the answer looks just a little simpler, but the steps ... lol, I'm not sure in the exam I'll figure out the steps. Probably the existing answer will do for me. – Jiew Meng Feb 26 '12 at 2:42

$y$ is assumed to be a function of $x$ here, it's not a constant. So your first solution is not correct (note WA interpreted the input as taking a partial derivative with respect to $x$, which is a different from what you want).

You need to use the "implicit differentiation" method to find the derivatives (you did this correctly in your second method):

$y$ is defined implicitly as a function of $x$ by the equation $$2x^3-3y^2=8.$$ Let's find the first derivative. To find $dy\over dx$, differentiate both sides of the above with respect to $x$ keeping in mind that $y$ is a function of $x$: $${d\over dx}(2x^3-3y^2)={d\over dx} 8.$$

$$6x^2-6y {dy\over dx}=0.$$ Note that we needed to use the chain rule to find ${d\over dx} 3y^2$.

Solving for ${dy\over dx}$ gives $${dy\over dx } ={x^2\over y }.$$

To find ${d^2y\over dx^2}$, differentiate both sides of the above with respect to $x$:

$${d\over dx}{dy\over dx } ={d\over dx}{x^2\over y }.$$ $${d^2y\over dx^2}= {2x y-{dy\over dx} x^2\over y^2} ;$$ simplifying the right hand of the above side leads to your solution.

Sometimes answers aren't simple; such is life...

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The reason WolframAlpha treats $y$ as a constant here is that as far as it knows, $y$ is constant with respect to $x$, and $\frac{d}{dx}$ denotes taking the derivative with respect to $x$. However, in your problem $y$ is a function of $x$ rather than an independent variable, so your approach is the correct one.

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$$2x^3-3y^2 = 8.$$ Differentiate both sides with respect to $x$: $$6x^2 - 6yy' = 0.$$ Solve for $y'$: $$y' = \frac{x^2}{y}.$$ Differentiate again with respect to $x$: $$y'' = \frac{y(2x)- x^2y'}{y^2}.$$ Put $x^2/y$ in place of $y'$ and simplify: $$y'' = \frac{2xy - x^2\frac{x^2}{y}}{y^2} = \frac{2xy^2 - x^4}{y^3}.$$

In a sense, you're done now, but notice that in place of $y^2$ you can put $(2x^3-8)/3$: $$y'' = \frac{2x(2x^3-8)/3 - x^4 }{y(2x^3-8)/3} = \frac{3x^4 - 16x}{y(2x^3-8)}.$$

Maybe depending on the purpose, you might prefer to put $(8+3y^2)/2$ in place of $x^3$: $$y'' = \frac{2xy^2-x(8+3y^2)/2}{y^3} = \frac{4xy^2-8x-3xy^2}{2y^3}.$$

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