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

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

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


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

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

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

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
up vote 2 down vote accepted

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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