# Finding the second derivative of a first derivative.

so I have a $\frac{ dy}{dx}=\frac{-2x+y}{2y-x}$ from the original equation $\ x^2+y^2-xy=1 \$.

I have to find the second derivative given a coordinate point $\left(\sqrt{\frac{1}{3}},2\sqrt{ \frac{1}{3}})\right)$. Here is my work:

$$\frac{d^2y}{dx^2} = \frac{(2y-x)(-2+y')-(-2x+y)(2y'-1)}{(2y-x)^2}$$

$$=\frac{(2y-x)\left(-2+\frac{-2x+y}{2y-x}\right)-(-2x+y)(\frac{-4x+2y}{2y-x}-1)}{(2y-x)^2}$$

then I started plugging in,

$$=\frac{\left(\frac{4}{\sqrt{3}}\right)-\frac{1}{\sqrt{3}}\left(-x+\frac{-2\sqrt{1/3}+2\sqrt{1/3}}{4\sqrt{1/3}-\sqrt{1/3}}\right)-[(-2\sqrt{1/3}+2\sqrt{1/3}\left(2[\frac{-2\sqrt{1/3}+2\sqrt{1/3}}{4\sqrt{1/3}-\sqrt{1/3}}]-1\right)]}{(4\sqrt{1/3}-\sqrt{1/3})^2}$$

Some of the expressions cancelled out to be 0, so I ended up with

$$\frac{d^2y}{dx^2}=\frac{3\sqrt{1/3}}{(3\sqrt{1/3})^2}=\frac{1}{\sqrt{3}}.$$ However, this wasn't the right answer. Is there an easier approach or something wrong with my calculation? Thank you.

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This is completely unintelligible! You need to edit this using latex. You can find the tutorial here. – homegrown May 19 '14 at 23:39
I have made a first cut at making this a bit more readable; I don't have time at the moment to pick through all the "sqrts" and such... – RecklessReckoner May 20 '14 at 2:33

I can't read the question, so I'll give the second derivative. Use the quotient rule: $$\dfrac{d^2y}{dx^2}=\dfrac{(2y-x)\left(-2+\frac{-2x+y}{2y-x}\right)-(-2x+y)\left(2\frac{-2x+y}{2y-x}-x\right)}{(2y-x)^2}$$ Substitute the given $(x,y)$ coordinate to find the second derivative there.
Remember $y$ depends on $x$, so you need to use implicit differentiation. – Grumpy Parsnip May 20 '14 at 2:39
Did you? I just assumed you applied the quotient rule straight, but I see you substituted in for $y'$. +1 – Grumpy Parsnip May 20 '14 at 4:39