Here's another:
If $\displaystyle ax^2 + by^2 = c$ show that the second derivative is equal to $\displaystyle -ac/b^2y^3$
Thanks
Because the second derivative will have an "x" in it and eliminating that reintroduces the "c".
Since $\displaystyle ax^2+ by^2= c$, $\displaystyle 2ax+ 2byy'= 0$. Differentiating again, $\displaystyle 2a+ 2by'^2+ 2byy"= 0$.
$\displaystyle y"= \frac{-a- by'^2}{by}$.
From $\displaystyle ax+ byy'= 0$, $\displaystyle y'= -\frac{ax}{by}$ and so $\displaystyle y'^2= \frac{a^2x^2}{b^2y^2}$.
Then $\displaystyle y"= -\frac{a+ \frac{a^2bx^2}{by^2}}{by}$
$\displaystyle y"= -\frac{aby^2+ a^2x^2}{b^2y^3}$
From $\displaystyle ax^2+ by^2= c$, $\displaystyle ax^2= c- by^2$ so $\displaystyle a^2x^2= ac- aby^2$.
$\displaystyle aby^2+ a^2x^2= aby^2+ ac- aby^2= ac$
$\displaystyle y"= -\frac{ac}{b^2y^3}$
1st derivative...
$\displaystyle \frac{d}{dx}(ax^2+by^2)=\frac{d}{dx}c$
$\displaystyle 2ax+b\frac{d}{dx}y^2=0$
$\displaystyle 2ax+b\frac{dy}{dx}\frac{d}{dy}y^2=0$
$\displaystyle 2ax+2yb\frac{dy}{dx}=0$
2nd derivative...
2by is one factor and y' is the other of 2by(y'), or y and 2by' etc.
$\displaystyle 2a+2by\frac{d}{dx}\frac{dy}{dx}+\frac{dy}{dx}2b\fr ac{dy}{dx}=0$
$\displaystyle 2a+2by\frac{d^2y}{dx^2}+2b(\frac{dy}{dx})^2=0$