# Second Derivative Doesn't Work!

1. Mar 16, 2005

### sjaguar13

Starting with y^2-xy=2, then y'=y/(2y-x)

y''=(2y-x)y'-y(2y'-1)/(2y-x)^2

substitute y' in:
y''=(2y-x)(y/2y-x) - y(2(y/2y-x)-1)

the (2y-x) cancels:
y''=(y) - y(2y/2y-x)-1)

distribute:
y'' = y - (2y^2)/(2y-x)-y

Somewhere I need y^2-xy so I can replace it with the original 2, but I'm not getting it.

2. Mar 16, 2005

### Data

No. He wants to implicitly differentiate to get $$\frac{d^2 y}{d x^2}$$.

$$y^2 - xy = 2 \Longrightarrow 2y\frac{d y}{dx} - y - x\frac{dy}{dx} = 0 \Longrightarrow \frac{dy}{dx} = \frac{y}{2y-x}$$

$$\Longrightarrow \frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}}{2y-x} \ - \ \frac{y(2\frac{dy}{dx} - 1)}{(2y-x)^2} = \frac{\frac{y}{2y-x}}{2y-x} \ - \ \frac{y\left( \frac{2y}{2y-x} - 1\right)}{(2y-x)^2}$$

$$= \frac{y}{(2y-x)^2} \ - \ \frac{y\frac{2y - (2y-x)}{2y-x}}{(2y-x)^2} = \frac{y}{(2y-x)^2} \ - \ \frac{yx}{(2y-x)^3} = \frac{y(2y-x) - xy}{(2y-x)^3} = \frac{2y^2 - 2xy}{(2y-x)^3} = \frac{4}{(2y-x)^3}$$

3. Mar 16, 2005

### sjaguar13

I don't really follow. Why is the second derivative = first deriv - (top)(bottom)/(bottom)^2?

In the book, the quotient rule is ((bottom)(top)' - (top)(bottom)')/(bottom)^2. That's what I tried to do, but it didn't work.

4. Mar 16, 2005

### arildno

Data chose to differentiate by using the product rule on the function $$y(x)*\frac{1}{2y-x}$$

And, I have to add, the second derivative ALWAYS works, even if you don't know what to use it for..

Last edited: Mar 16, 2005
5. Mar 16, 2005

### Hurkyl

Staff Emeritus
Instead of looking for a y^2 - xy that you can replace with 2, maybe you could figure out what to replace y^2 with? Why'd you want to do that anyways?

(I'm assuming your work was right)

6. Mar 16, 2005

### SpaceTiger

Staff Emeritus
That's not what you did. Look above. It should be y'/(2y-x)-y(2y'-1)/(2y-x)^2.