Question

When asked to find d2y/dx2 implicitly in terms of x and y for x^2-y^2=9, I keep getting (y^2-x^2)/y^2 for my y'' and it should be (x^2-y^2)/y^2. What am I doing wrong?!

Answer 1

lets check,
\(\Large 2x-2yy' =0 \\ \large y' = \dfrac{x}{y}\)

Answer 2

\(\Large y'' = \dfrac{y-xy'}{y^2}\)

but \(y' =\dfrac{x}{y} \)

so,
\(\Large y'' = \dfrac{y-x(x/y)}{y^2} = \dfrac{y^2-x^2}{y^3}\)

why do u sat the answer should be (x^2-y^2)/y^2. ??

Answer 3

*say

Answer 4

alternate way,
\(\Large x=yy' \\ \large 1 = y'^2 +yy'' \\ \Large 1= x^2/y^2 +yy'' \\ \Large 1-x^2/y^2 = yy'' \\ \large y'' = \dfrac{y^2-x^2}{y^3}\)
getting the same answer again!, so it must be correct :P

Answer 5

because the original function was x^2-y^2 in the numerator. For the problem to work implicitly so that I can substitute 9 back into my y'', the numerator must equal x^2-y^2

Answer 6

why *must* ??
if the numerator is y^2-x^2, we can always say that since,

x^2-y^2 = 9
y^2-x^2 = -9

and our y" = -9/y^3

Answer 7

Okay, I just wasn't sure if I could do that, but it is what I wanted to do all along, thanks

Answer 8

all doubts cleared ? anything left anywhere ?

Answer 9

No, we just never manipulated the original function when substituting before. I wasn't sure if it was safe to do. I'm good now.

Answer 10

welcome ^_^

Answer 11

Thanks again