Implicit differentiation, find y''

[apiwowar]

the problem is to find y'' or d²y / d²x

the equation is y² = x²

first i found the first derivative dy/dx = 2x / 2y = x / y

then i found the second using the quotient rule and got

y'' = (y - x(dy/dx)) / y²

i plugged in y' into y'' and got

y'' = (y - (x²/y)) / y²

but then I am stuck after that because when i simplify i get
y'' = (y² - x²) / y³

i don't know how i can plug the original equation back in.

did i go wrong somewhere?

 

[Mark44]

the problem is to find y'' or d²y / d²x

the equation is y² = x²

first i found the first derivative dy/dx = 2x / 2y = x / y

This looks fine.

then i found the second using the quotient rule and got

y'' = (y - x(dy/dx)) / y²


i plugged in y' into y'' and got

y'' = (y - (x²/y)) / y²

but then I am stuck after that because when i simplify i get
y'' = (y² - x²) / y³

This looks fine, too.

i don't know how i can plug the original equation back in.

This above is correct, but can be simplified. If y² = x², then y² - x² = ?

did i go wrong somewhere?

 

[ehild]

Take care, you have to exclude x=y=0.
The original equation is y²=x². What do you get when you plug that in?

ehild

 

[apiwowar]

does x²-y² = 0 or 1?

and the book shows that the answer is 3y/4x² = 3x / 4y, hows that possible? shouldn't the answer be y'' = ...?

 

[Mark44]

does x²-y² = 0 or 1?

If two numbers are equal, then subtracting one from the other leaves zero.

and the book shows that the answer is 3y/4x² = 3x / 4y, hows that possible? shouldn't the answer be y'' = ...?

Yes, they're asking you for y'', so the answer should start with y'' = ...

I don't see how they got 3y/4x² = 3x / 4y.

Going back to the original problem, you have y² = x², which is equivalent to two equations: y = x or y = -x. For the first, y' = 1, and for the second, y' = -1. What do these tell you about y''?