1. ## Differentials

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.

2. Originally Posted by Cursed
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
$\displaystyle x^2y + xy^2 = 6$

$\displaystyle 2xy + 2x \frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$

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

3. 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:

$\displaystyle y''=\frac{2(y-x(\frac{-2y}{x}))}{x^{2}}=\frac{6y}{x^{2}}$