# Homework Help: D2y / dx2

1. Feb 27, 2010

### quietrain

d2y/dx2 = d/dy [1/2 (y'^2)]

so how do we show that the LHS = RHS? starting from the LHS?

i don't really understand.

my lecturer made it become d2y/dx2 = dy/dx * d/dy * y' and he showed it equals RHS

i understand how he get to dy/dx * d/dy * y'

but shouldn't that become dy/dx *y' * d/dy which is (y')^2 * d/dy? so where did the 1/2 from the RHS come from?

any help? thanks!

2. Feb 27, 2010

### tiny-tim

hi quietrain!

(try using the X2 tag just above the Reply box )
(first, never write " d/dy * ", it makes no sense, either write " d/dy(y') " or " dy'/dy " )

now convert dy/dx to to y', and it's y' * d/dy (y'), = d/dy (1/2 y'2)

3. Feb 28, 2010

### quietrain

oh i see.

so d2y / dx2 = d/dx dy/dx = d/dy dy/dx *dy/dx = y' * d/dy y'

so i integrate y' * d/dy y' = 1/2 y'2

so the differiential form of 1/2 y'2 = y' *d/dy y'

but i don't understand this, what is d/dy y'?

i know d/dx of y' = d/dx of dy/dx which means differientiate dy/dx one more time with respect to x

but if i have d/dy of y' then i am differientiating dy/dx with respect to y ?

so for example y = x3

dy/dx = 3x2

so d/dx dy/dx = 6x

but what is d/dy dy/dx ?

4. Feb 28, 2010

### tiny-tim

d/dy dy/dx = d/d(x3) 3x2

= d/dx3 3[x3]2/3

= 3*(2/3)*[x3]-1/3

= 2x-1

so y' d/dy dy/dx = 3x2 2x-1 = 6x

5. Mar 1, 2010

### quietrain

oh isee...

so thats why we need to put in a y' for partial differientiation...

so we have to convert the x2 term into a x3 term so that we can differientiate.

thanks a lot tim!!!