Question

Ok I am a little stuck. Let x^3+y^3=65. Find y"(x) at the point (4,1). I found Y(') but cant seem to get the answer for the next double prime.

Answer 1

I think y(') is \[\frac{ -x^2 }{ y^2 }\]

Answer 2

I got that as well.

Answer 3

Are you familiar with quotient rule?

Answer 4

Yea thats how I set it up next but I must have made a mistake

Answer 5

I had (-2xy^2+x^2(2y)dy/dx)/y^4

Answer 6

On the top, when you take derivative of y^2 should get 2yy' :)

Answer 7

oh thats what I meant where the dy/dx is

Answer 8

man, sorry. That's right!

Answer 9

I dont use y' b/c I always make a mistake lol

Answer 10

I think my problem is in simple algebra that I can get the y' alone :(

Answer 11

don't need to. Substitute \( \large \dfrac{dy}{dx} = \dfrac{-x^2}{y^2}\)

Answer 12

Oh

Answer 13

I forgot part of the problem is says y"(4)= is what its asking

Answer 14

Oh, good. So you just need to substitute x = 4, y = 1, and y' = -16

Answer 15

where did u get 16?

Answer 16

y'(4) = -(4)^2/(1)^2

Answer 17

ahh

Answer 18

y''(4) = -520?

Answer 19

Yeah ty!

Answer 20

np!

Answer 21

I only have 4 more left :P this homework is so long lol