Question

Find d2y/dx2 in terms of x and y if (x^2)-(y^2)=16

Answer 1

differentiate implicitly and solve for dy/dx
show what you get

Answer 2

i got for y'=2x-2y and for y''=0

Answer 3

that just doesnt look right to me

Answer 4

d/dx(x^2)+d/dx(y^2)=d/dx(5)
2x+2y(dy/dx)=0
dy/dx=-x/y

Answer 5

now differentiate implicitly again
whenever you get dy/dx in what you are doing you can sub in -x/y

Answer 6

how did you get 5

Answer 7

or 16 whatever
it's a constant so either way it is zero upon taking the derivative

Answer 8

d/dx(x^2)+d/dx(y^2)=d/dx(16)
2x+2y(dy/dx)=0
dy/dx=-x/y

Answer 9

doesn't change anything

Answer 10

ok i see it now

Answer 11

so how do i get the second prime from that? i dont see

Answer 12

d/dx(dy/dx)=d/dx(-x/y)
you can use the quotient rule

Answer 13

ok i got -16/y^3

Answer 14

where did -16 come from?

Answer 15

idk thats what i got when i put it in wolfram look
http://www.wolframalpha.com/input/?i=%28x%5E2%29-%28y%5E2%29%3D16

Answer 16

I see nothing about the second derivative in that link

Answer 17

you have to press more next to implicit derivatives

Answer 18

oh ic

Answer 19

well it might help to derive that, plus I would imagine they want that in terms of x anyway

Answer 20

sorry im just really bad with derivatives

Answer 21

?

Answer 22

\[x^2+y^2=16\]\[2x+2yy'=0\implies y'=-\frac xy\]\[y''=-\frac d{dx}\left(\frac xy\right)=-{y\frac d{dx}(x)-x\frac d{dx}(y)\over y^2}\]\[y''=-{y-x(-\frac xy)\over y^2}=-\frac{y^2+x^2}{y^3}=-\frac{16}{y^3}\]

Answer 23

now i see it !!! is that chain rule?

Answer 24

yes, exactly
chain rule on y(x)
d/dx(f(y))=f'(y)(dy/dx) <- chain rule

Answer 25

i didn't know that you could use chain rule with fractions that way

Answer 26

the quotient rule is really just the product rule + chain rule:\[(\frac uv)'=(uv^{-1})'=u'v^{-1}-uv^{-2}v'=\frac{u'v-v'u}{v^2}\]

Answer 27

i see!! thank you so much