does d^2y / dx^2 mean second derivative?
and how do u take the derivative of something like this
3y^2 - 4x=16 ?
Yes
i suppose the derivative you are talking about is dy/dx, you can find this two ways, explicitly or implicity.
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)
Hello, juiicycouture!
I'll assume you want the second derivative . . .How do u take the derivative of something like this? .3y² - 4x .= .16
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)