# Find the second derivative of some implicit function?

I have a function given implicitly, you know. X og Y on both sides. Than it says, assume y=y(x). That's fine. And find y'(0), that I should be able to do. But what about y''(0)? How do you treat the dy/dx parts when taking the second derivative?

Edit: I would also like to follow the tip in the book, that says when I'm after actual values. We can just insert the value instead solving for dy/dx.

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You simply replace all the $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms in your second derivative with the expression you got for $\frac{\mathrm{d}y}{\mathrm{d}x}$ through implicit differentiation.

As an explicit example, suppose we wanted to find $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ for the expression $x^2+y^2-r=0$ ($r$ here is a constant). We find

$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}$$

and differentiating again gives

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

If we substitute the expression we got for the first derivative into the second derivative, we get

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

which gives

$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{x^2+y^2}{y^3}$$

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First you find $dy/dx$, say $$\frac{dy}{dx}=g(x,y).$$ Then by chain rule $$\frac{d^2y}{dx^2}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\frac{dy}{dx}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}g(x,y).$$