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I am trying to find the second order derivative of following implicit function.

$$y^3-2xy+4=0$$

I can find the first order no problem. For $y(x)$ this gives:

$$(3y^2-2x)y'=2y \Rightarrow y'=\frac{2y}{3y^2-2x}$$

Halfway through the second order derivative however, I seem to get lost and end up with something different compared to the solution.

I get:

$$(3y^2-2x)y''=2y'-(6yy'-2)y'$$

Which provides a completely different answer compared to the solution that my book gives at this point, which is the following:

$$(3y^2-2x)y''=(4-6yy')y'$$

The change in signs i can understand, but they seem to be multiplying the $2$ from the $2y'$ in the right part of the function with the $2$ inside the parenthesis of $(2-6yy')$. Which does not make much sense to me as there is no multiplication going on between these two terms, and even if there was then I would think it would result in $(4-12yy')$ instead, no?

Any pointers are appreciated.

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    $\begingroup$ "They seem to be multiplying" but they are not, they are infact subtracting $6yy^{'}-2$ from $2$. It is a coincidence that $2+2 = 2 \times 2$. $\endgroup$ Dec 28, 2016 at 12:49

2 Answers 2

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\begin{align} (3y²-2x)y'' & =2y'-(6yy'-2)y'\\ (3y²-2x)y'' & =2y'-6yy' y'+2y'\\ (3y²-2x)y'' & =4y'-6yy' y'\\ (3y²-2x)y'' & =(4-6yy') y'\\ \end{align}

Clear?

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  • $\begingroup$ Clear! I can't believe I overlooked that. Thank you. $\endgroup$ Dec 28, 2016 at 13:11
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You always have that $$ 2a-(6b-2)a=4a-6ab=(4-6b)a. $$

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