Question

find the second derivative of 2x^2-y^2=1 using implicit differentiation

Answer 1

start with
\[4x-2yy'=0\] then solve for \(y'\)

Answer 2

Okay that gave me 2x/y and that's where I get lost

Answer 3

looks good to me

Answer 4

unless you solve for \(y\) explicitly in terms of \(x\) your answer will have both an \(x\) and a \(y\) in it

Answer 5

I understand that I 'm trying to find teh second derivative.

Answer 6

oooh second derivative!! my fault, sorry
now you have to take the derivative of \(\frac{2x}{y}\) using the quotient rule

Answer 7

\[\frac{2y-2xy'}{y^2}\] is the first step via the quotient rule

Answer 8

I got that, I just don't know where to go from there!

Answer 9

second step is to replace \(y'\) by \(\frac{2x}{y}\) and then some algebra

Answer 10

\[\frac{2y-2xy'}{y^2}\]
\[\frac{2y-2x\frac{2x}{y}}{y^2}\]

Answer 11

so, 2y-(2x^2/y)/y^2

Answer 12

multiply top and bottom by \(y\) to clear the fraction, get
\[\frac{2y^2-4x^2}{y^3}\]

Answer 13

OH!!

Answer 14

yeah you are right, need to clean it up a little that's all

Answer 15

we can clean it up even more

Answer 16

\[2y^2-4x^2=-2(2x^2-y^2)=-2\times 1\] pretty tricky huh?

Answer 17

Okay, thanks for the quick response and help!! My answer key doesn't want the simplified version, but neat trick!

Answer 18

yw