Find the second derivative of some implicit function?
I have a function given implicitly, you know. X and Y on both sides. Then it says, assume y = y(x). That's fine. I should be able to find y'(0), 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.
$\endgroup$2 Answers
$\begingroup$If you can use partial derivatives, then you can do the following:
First you find $dy/dx$, say $$\frac{dy}{dx}=g(x,y).$$ Then by chain rule $$\frac{d^2y}{dx^2}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\frac{dy}{dx}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}g(x,y).$$
$\endgroup$ $\begingroup$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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