Proving LHS = RHS: d2y/dx2 and 1/2(y'^2)

[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!

 

[tiny-tim]

hi quietrain!

(try using the X² tag just above the Reply box )

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' …

(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)

 

[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 ?

 

[tiny-tim]

… 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 ?

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

 

[quietrain]

oh isee...

so that's 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!