Question

If x^4+y^4=1, which expression for y" ( the 2 nd derivative)

Answer 1

Do you know how to find the first derivate ?

Answer 2

I believe it's 4x^3+4yy'=0

Answer 3

Close . it's actually 4x^3+4y^3y'=0

Answer 4

Can you now solve it with respect to y' ?

Answer 5

(-4x^3)/(4y^3)

Answer 6

now diferentiate one more time

Answer 7

this expression (-4x^3)/(4y^3)

Answer 8

-x^3/y^3

Answer 9

hmm something aint right with this one

Answer 10

By applying the quotient rule I get this (-12x^2)(4y^3) - (-4x^3)(y’12y^2) / (4y^3)^2

Answer 11

[ (-12x^2)(4y^3) - (-4x^3)(y’12y^2) ] / (4y^3)^2

Answer 12

now the trick here is to notice that we have another y' since we differentated again

Answer 13

We already know what y' equals to since this is the first derivate that we found before

Answer 14

so you need to replace it likeso [ (-12x^2)(4y^3) - (-4x^3)(((-4x^3)/(4y^3))12y^2) ] / (4y^3)^2

Answer 15

https://www.dropbox.com/s/j5wkqlfy251bnux/Screenshot%202016-01-13%2022.09.53.png?dl=0

Answer 16

Do you know what is the quotient rule ?

Answer 17

any questions ? :o

Answer 18

Yeah

Answer 19

So basically you need to remember to make this substitution on the second derivative

Answer 20

We found that y' = (-4x^3)/(4y^3)

Answer 21

And then that y'' = [ (-12x^2)(4y^3) - (-4x^3)(y’12y^2) ] / (4y^3)^2

Answer 22

y'' = [ (-12x^2)(4y^3) - (-4x^3)( -------- > y’ <------- 12y^2) ] / (4y^3)^2 This needs to be replaced with (-4x^3)/(4y^3)

Answer 23

Thx