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

and how do u take the derivative of something like this

3y^2 - 4x=16 ?

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- Apr 5th 2007, 10:23 PMjuiicycouturedoes 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, 10:36 PMJhevon
Yes

Quote:

and how do u take the derivative of something like this

3y^2 - 4x=16 ?

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, 11:40 PMearboth
- Apr 5th 2007, 11:43 PMJhevon
- Apr 6th 2007, 09:10 AMSoroban
Hello, juiicycouture!

Quote:

How do u take the derivative of something like this? .3y² - 4x .= .16

__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)**