# does d^2y / dx^2 mean second derivative?

• Apr 5th 2007, 09:23 PM
juiicycouture
does d^2y / dx^2 mean second derivative?
does d^2y / dx^2 mean second derivative?

and how do u take the derivative of something like this

3y^2 - 4x=16 ?
• Apr 5th 2007, 09:36 PM
Jhevon
Quote:

Originally Posted by juiicycouture
does d^2y / dx^2 mean second derivative?

Yes

Quote:

and how do u take the derivative of something like this

3y^2 - 4x=16 ?
i suppose the derivative you are talking about is dy/dx, you can find this two ways, explicitly or implicity.

implicity:

3y^2 - 4x = 16
=> 6y y' - 4 = 0
=> y' = 4/6y = 2/3y
and if you want, you can solve for y and plug it in

explicitly:
we solve for y first and then take the derivative normally

3y^2 - 4x = 16
=> 3y^2 = 16 + 4x
=> y^2 = (16 + 4x)/3
=> y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) = (16/3)^(1/2) + (4x/3)^(1/2) .................incorrect, see post below
=> y' = (1/2)(4x/3)^(-1/2) * (4/3) = (2/3)(4x/3)^(-1/2)
• Apr 5th 2007, 10:40 PM
earboth
Quote:

Originally Posted by Jhevon
...
explicitly:
we solve for y first and then take the derivative normally

3y^2 - 4x = 16
=> 3y^2 = 16 + 4x
=> y^2 = (16 + 4x)/3
=> y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) = (16/3)^(1/2) + (4x/3)^(1/2)
=> y' = (1/2)(4x/3)^(-1/2) * (4/3) = (2/3)(4x/3)^(-1/2)

Hello, Jhevon,

I will not comment this kind of transformation - but it is extremly wrong:

Repair:

y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) ===>

y' = ½ * ((1/3)(16+4x))^(-½ )*4 = 2/(sqrt((1/3)(16+4x))

EB
• Apr 5th 2007, 10:43 PM
Jhevon
Quote:

Originally Posted by earboth
Hello, Jhevon,

I will not comment this kind of transformation - but it is extremly wrong:

Repair:

y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) ===>

y' = ½ * ((1/3)(16+4x))^(-½ )*4 = 2/(sqrt((1/3)(16+4x))

EB

Ah yes, i see my mistake. my bad, i'm half asleep right now
• Apr 6th 2007, 08:10 AM
Soroban
Hello, juiicycouture!

Quote:

How do u take the derivative of something like this? .3y² - 4x .= .16
I'll assume you want the second derivative . . .

Differentiate implicitly: .6y(dy/dx) - 4 .= .0

. . Then: .dy/dx .= .2/(3y) .= .(2/3)y^
(-1) .[1]

Differentiate again: .d²y/dx² .= .-(2/3)y^
(-2) (dy/dx)

Substitute [1]: .d²y/dx² .= .-(2/3)y^
(-2) · (2/3)y^(-1)

. . Therefore: . d²y/dx² .= .-4/(9y³)