# Implicit differentiation, find y''

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 im stuck after that because when i simplify i get
y'' = (y2 - x2) / y3

i dont know how i can plug the original equation back in.

did i go wrong somewhere?

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Mark44
Mentor
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 im stuck after that because when i simplify i get
y'' = (y2 - x2) / y3
This looks fine, too.
i dont 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
Homework Helper
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

does x2-y2 = 0 or 1?

and the book shows that the answer is 3y/4x2 = 3x / 4y, hows that possible? shouldnt the answer be y'' = ......?

Mark44
Mentor
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? shouldnt the answer be y'' = ......?