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What would the 2nd order implicit derivative of $y^2=12x$ be?

I get the first derivative is $2yy'=12$ but Wolfram gives the second as $y''=\frac{-3}{xy}$ and I don't understand how they get that.

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The first derivative you obtained is correct, $$\tag{1} 2 y y' = 12.$$

Taking again a derivative with respect to $x$, we obtain $$ \tag{2} 2 y'^2 +2 y y'' =0.$$

Eliminating $y'$ from (1) and (2) and solving for $y''$ yields $$ y'' = - \frac{36}{y^3}.$$

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$$ y^2 = 12x \Rightarrow \\ 2\cdot y \frac{dy}{dx} = 12 \Rightarrow \\ \frac{dy}{dx} = \frac{12}{2y} = \frac{6}{y} \Rightarrow \\ \frac{d^2y}{dx^2} = -\frac{6}{1}\cdot \frac{1}{y^2}\cdot \frac{dy}{dx} = -\frac{6}{1}\cdot \frac{1}{y^2}\cdot \frac{6}{y} = -\frac{36}{y^3} = -36\cdot \frac{1}{y^2\cdot y} = -36\cdot \frac{1}{12xy} = \frac{-3}{xy} $$

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Differentiating both sides of $y(x)y'(x)=6$, we have $$\frac{d}{dx}(y(x)y'(x))=\frac{d}{dx}(6), \\ (y'(x))^2+y(x)y''(x)=0, \\ $$ thus $$y''(x)=-\frac{(y'(x))^2}{y(x)}=-\frac {36}{y^3}=-\frac {36}{12xy}=-\frac{3}{xy}.$$

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Thank you everyone! It wasn't so complicated after all: I only had to use the product rule to start it. – Ian Feb 14 '13 at 9:42

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