If $\displaystyle x^2y + yx^2 = 6$, then what does $\displaystyle \frac{d^2y}{dx^2}$ equal at the point $\displaystyle (1,3)$?
**I have not attempted this solution because I do not understand how $\displaystyle \frac{d^2y}{dx^2}$ works.
I am going to assume this was supposed to be
$\displaystyle x^2y + xy^2 = 6$
Start with your first derivative:
$\displaystyle 2xy + 2x \frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$
so
$\displaystyle \frac{dy}{dx} = -\frac{2xy + y^2}{2x + 2xy}$
Now take the second derivative:
$\displaystyle \frac{d^2y}{dx^2} = -\frac{\left ( 2y + 2x\frac{dy}{dx} \right ) (2x + 2xy) + (2xy + y^2) \left ( 2 + 2y + 2x \frac{dy}{dx} \right ) }{(2x + 2xy)^2}$
Now plug in your expression for the derivative and simplify.
-Dan
Well, we can differentiate implicitly and then do it again.
$\displaystyle x^{2}y+x^{2}y=6$
$\displaystyle 2(x^{2}y)=6$
$\displaystyle x^{2}y=3$
Product rule:
$\displaystyle x^{2}y'+2xy=0$
$\displaystyle y'=\frac{-2xy}{x^{2}}=\frac{-2y}{x}$
Do it again using quotient rule:
$\displaystyle y''=\frac{-2xy'+2y}{x^{2}}=\frac{2(y-xy')}{x^{2}}$
But, note what y' is equal to from before. Sub that in:
$\displaystyle y''=\frac{2(y-x(\frac{-2y}{x}))}{x^{2}}=\frac{6y}{x^{2}}$
Plug in your values.