x^-1 +y^-1 = pi^-1
How do I differentiate this? dy/dx should be -(y^2)/x^2.
Hello, Stuck Man!
$\displaystyle \text{Is the second derivative: }\:y'' \:=\:\frac{2y^3 +2xy^2}{x^4}\:?$
Yes . . . but it can be further simplified.
We have: .$\displaystyle y'' \;=\;\frac{2y^2(y + x)}{x^4}$ .[1]
From the original equation: .$\displaystyle \frac{1}{x} + \frac{1}{y} \:=\:\frac{1}{\pi}$
. . we have: .$\displaystyle \frac{y+x}{xy} \:=\:\frac{1}{\pi} \quad\Rightarrow\quad y+x \:=\:\frac{xy}{\pi}$
Substitute into [1]: .$\displaystyle y'' \;=\;\frac{2y^2\left(\frac{xy}{\pi}\right)}{x^4} \;=\;\frac{2y^3}{\pi x^3}$
This is something to watch for . . .
Quite often, the second derivative contains part (or all) of the original equation.