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Second order partial differential of implicite function

Posted December 3rd, 2009 by Structure

Let $G(x, y, z)=F(\frac{x}{z},\frac{y}{z})=0$ an equation defined by a function $F\in \mathcal{C}^2$
Suppose we can apply the implicit function theorem in a neighborhood of a given point (a,b,c).
So we suppose

$G(a,b,c)=F(\frac{a}{c},\frac{b}{c})=0$

and

$\frac{\partial G}{\partial z}(a,b,c)=\frac{\partial F}{\partial u}(\frac{a}{c},\frac{b}{c})(-\frac{a}{c^2})+\frac{\partial F}{\partial v}(\frac{a}{c},\frac{b}{c})(-\frac{b}{c^2})\ne 0$

Then there is a function z=f(x,y) with f(a,b)=c; G(x,y,f(x,y))=0 or

$F(\frac{x}{f(x,y)},\frac{y}{f(x,y)})=0$

We have

List of antiderivatives

Posted September 29th, 2009 by Structure

$\int x^ndx=\frac{x^{n+1}}{n+1}+\Math{C}\;;\int f^n(x)f'(x)dx=\frac{f^{n+1}(x)}{n+1}+\Math{C}$

$\int \frac{1}{x}dx=\ln|x|+\Math{C}\;;\int \frac{f'(x)}{f(x)}dx=\ln|f(x)|+\Math{C}$

$\int \sin xdx=-\cos x+\Math{C}\;;\int \sin f(x)f'(x)dx=-\cos f(x)+\Math{C}$

$\int \frac{1}{\sin x}dx=\int \frac{1}{2\sin \frac{x}{2}\cos\frac{x}{2}}dx=\int \frac{1}{2\tan\frac{x}{2}\cos^2\frac{x}{2}}dx=\ln|\tan\frac{x}{2}|+\Math{C}$

$\int \cos xdx=\sin x +\Math{C}\;;\int \cos f(x)f'(x)dx=\sin f(x)+\Math{C}$

Circumcircles of orthogonal four points

Posted September 9th, 2009 by Structure

Second order partial differential of implicite function

List of antiderivatives

Circumcircles of orthogonal four points

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