Question

find the second derivative of x^2-y^2=36 in terms of x and y

Answer 1

if you want in terms of x and y you want to do implicit differentiation:
\[x^2 - y^2 = 36\]
\[2xdx - 2ydy = 0\]
\[2dx^2 - 2dy^2 = 0\]

Answer 2

yea i know dy/dx=x/y but idk how to take the second derivative of it

Answer 3

this is the total derivative, given by:
\[f^\prime=\frac{\partial}{\partial{x}}+\frac{\partial}{\partial{y}}\]
\[f^{\prime\prime}=\frac{\partial^2}{\partial{x}}+2\frac{\partial^2}{\partial{x}\partial{y}}+\frac{\partial^2}{\partial{y}}\]

Answer 4

I am assuming that is what they want?

Answer 5

they want the second derivative

Answer 6

if they want d^2y that is different, this assumes both variables are independent.

Answer 7

yes thats what they want

Answer 8

so it should just be
\[2dx^2 - 2dy^2 = 0\]

Answer 9

idk it says the answer is -36/y^3

Answer 10

ok, so they aren't independent then.. I was just thinking that they weren't cause they aren't wrting
\[f(x,y) = x^2 - y^2\]

Answer 11

yes like that

Answer 12

no x, that's odd... gonna hit the whiteboard a sec.

Answer 13

\[\Large x^2+y^2=36\]\[\Large 2x+2yy'=0\quad\to\quad y'=\frac{x}{y}\]Then to find the second derivative we apply the quotient rule:\[\Large y''=\frac{y-xy'}{y^2}\]And then to simplify it down to the answer key I guess we need to plug some stuff in.. hmm

Do you understand how I applied quotient rule there?

Answer 14

no the equation is x2-y2=36

Answer 15

Ahh sorry typo :(
The first derivative should be correct though.
I meant to write x^2-y^2.

Answer 16

But did you understand the quotient rule process? :U
I know it can be a little tricky.

Answer 17

ah I see yes

Answer 18

yes i understand, so it will be \[\frac{ y-x(\frac{ x }{ y}) }{ y ^{2}}\]

Answer 19

Looks good so far!

Answer 20

idk what comes next

Answer 21

right so you get:
\[\frac{y^2 - x^2}{y} * \frac{1}{y^2}\]
\[\frac{y^2-x^2}{y^3}\]
which is the negative of the original equation.
\[\frac{-36}{y^3}\]

Answer 22

how did you get y^2-x^2?

Answer 23

because:
\[\frac{y - x(\frac{x}{y}))}{y^2}\]
\[\frac{\frac{y^2}{y} - \frac{x^2}{y}}{y^2}\]
\[\frac{\frac{y^2 - x^2}{y}}{y^2}\]
\[\frac{y^2 - x^2}{y}*\frac{1}{y^2}\]

Answer 24

i understand why x^2 but idk how you get y^2?

Answer 25

because I'm multiplying by y/y to get a common denominator

Answer 26

oooh okay got it!

Answer 27

\[1 * y = \frac{y}{y}*\frac{y}{1} = \frac{y^2}{y}\]

Answer 28

cool, good question... I learned quite a bit as well :)

Answer 29

thank you so much!

Answer 30

no problem, good @zepdrix stepped in though :)