# How to find $\frac{d^2x}{dy^2}$

How to find $\dfrac{d^2x}{dy^2}$ in terms of $\dfrac{d^2y}{dx^2}$ and $\dfrac{dy}{dx}$ for any implicit function which is twice differentiable w.r.t. both x and y.

I tried but we can't write $\dfrac{d^2x}{dy^2}$ as $\dfrac{d(dx)}{d(dy)}$ separately ... can we?

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I think this is what you were asking for

$\dfrac {dx}{dy}=\dfrac {1}{dy/dx}$

$\dfrac {d²x}{dy²}=\dfrac {d}{dy} \dfrac {1}{dy/dx}$

$\dfrac {d²x}{dy²}=-\dfrac {d}{dy} \dfrac {dy/dx}{(dy/dx)²}$

$\dfrac {d²x}{dy²}=- \dfrac {d}{dx} \dfrac{dy}{dx} \dfrac {\dfrac {dx}{dy}}{(dy/dx)²}$

$\dfrac {d²x}{dy²}=-\dfrac {d²y}{dx²}.\dfrac {1}{dy/dx} \dfrac {1}{(dy/dx)²}$

$\dfrac {d²x}{dy²}=-\dfrac {d²y}{dx²} . \dfrac {1}{(dy/dx)^3}$

$\dfrac {d²x}{dy²}=-\dfrac {d²y/dx²}{(dy/dx)^3}$

Is this the relation that you wanted ?

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Hmm.. YES! (+1) it's what was needed . But can you use some \dfrac in your LaTeX to make it more appealing to eyes..... – ABC Apr 1 '13 at 11:55
Should learn how to use \dfrac. Will keep that in mind though – lsp Apr 1 '13 at 11:57
@learner I think even you din't get those two lines and so didn't added \dfrac there . lol – ABC Apr 1 '13 at 12:43