Hard? implicit differentiation

[rook_b]

Homework Statement

Find d^2y/dx^2 in terms of x and y.
$$x^2y^2-2x=3$$

Homework Equations

property rule, chain rule, quotient rule,

The Attempt at a Solution

I can do this the long way, but there must be a shorter solution. Can I simplify it? I've found $$dy/dx=(-xy^2 +2x)/(2x^2y)$$. but the derivative of that takes a ridiculous amount of steps. If there isn't a shorter way then I'll just take my time.

 

[rocomath]

this is how i would simplify

$$(xy)^{2}-2x=3$$

there aren't that many steps, it only took me 3 steps, maybe you're doing something wrong, if you want show your steps and i'll be happy to help further.

 

[rook_b]

OK thanks, that simplification made the first derivative easier and pointed out to me that my original calculation had a mistake. My math skills have quite a few holes. So now I have $$dy/dx= (-xy^2+1)/(x^2y)$$ so is there another simplification I'm overlooking? If I differentiate that it's still a bit messy and I'm learning there is always an easier way... well sometimes.

 

[cristo]

Your answer for dy/dx in post #3 is correct. I'm not sure how you want to simplify this. You could write $$\frac{1-xy^2}{x^2y}=\frac{1}{x^2y}-\frac{y}{x}$$ but I'm not sure that this is simpler to differentiate.

 

[HallsofIvy]

I'm wondering what "long way" you used. Straight forward implicit differentiation does seem that long to me!

$$x^2y^2- 2x= 3$$
Differentiating once:
$$2xy^2+ 2x^2yy'- 2= 0$$
Differenitiating again:
$$2y^2+ 4xyy'+ 4xyy'+ 2x^2(y')^2+ 2x^2yy"= 0$$
$$-2x^2yy"= 2y^2+ 8xyy'+ 2x^2(y')^2$$
Now divide by $-2x^2y$ to get
$$-\frac{2y^2+ 8xyy'+ 2x^2(y')^2}{2x^2y}$$

Since the derivative will necessarily have "y" in it, I would see no reason to write y' as a function of x which I think is the "hard" part of what you did.