Differentials

**Cursed** · April 17th 2008, 01:58 PM

If $x^2y + yx^2 = 6$ , then what does $\frac{d^2y}{dx^2}$ equal at the point $(1,3)$ ?

**I have not attempted this solution because I do not understand how $\frac{d^2y}{dx^2}$ works.

**topsquark** · April 17th 2008, 02:11 PM

Originally Posted by Cursed

If $x^2y + yx^2 = 6$ , then what does $\frac{d^2y}{dx^2}$ equal at the point $(1,3)$ ?

**I have not attempted this solution because I do not understand how $\frac{d^2y}{dx^2}$ works.

I am going to assume this was supposed to be
$x^2y + xy^2 = 6$

Start with your first derivative:
$2xy + 2x \frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$

so
$\frac{dy}{dx} = -\frac{2xy + y^2}{2x + 2xy}$

Now take the second derivative:
$\frac{d^2y}{dx^2} = -\frac{\left ( 2y + 2x\frac{dy}{dx} \right ) (2x + 2xy) + (2xy + y^2) \left ( 2 + 2y + 2x \frac{dy}{dx} \right ) }{(2x + 2xy)^2}$

Now plug in your expression for the derivative and simplify.

-Dan

**galactus** · April 17th 2008, 02:19 PM

Well, we can differentiate implicitly and then do it again.

$x^{2}y+x^{2}y=6$

$2(x^{2}y)=6$

$x^{2}y=3$

Product rule:

$x^{2}y'+2xy=0$

$y'=\frac{-2xy}{x^{2}}=\frac{-2y}{x}$

Do it again using quotient rule:

$y''=\frac{-2xy'+2y}{x^{2}}=\frac{2(y-xy')}{x^{2}}$

But, note what y' is equal to from before. Sub that in:

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

Plug in your values.

**Cursed** · April 17th 2008, 02:25 PM

Thank you both for the help. I understand it now.

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