d2y/dx2 at point (-2,1)

• Sep 10th 2009, 07:58 PM
genlovesmusic09
d2y/dx2 at point (-2,1)
Find d2y/dx2 at point (-2,1) for y^2-2x-4y-1=0

so i tired moving the xs and constants over:
y^2-4y=2x+1
but then I'm stuck...
• Sep 11th 2009, 12:26 AM
chisigma
The explicit expression for $\displaystyle y(x)$ is found resolving the algebraic equation...

$\displaystyle y^{2} - 4 y - 2x -1 =0$ (1)

... and we obtain...

$\displaystyle y = 2 \pm \sqrt {2x+5}$ (2)

Because is $\displaystyle y(-2) =1$ we have to choose in (2) the sign '-', so that is...

$\displaystyle \frac{d^{2} y}{dx^{2}} = (2x+5)^{- \frac{3}{2}} = 1 _{x=-2}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 11th 2009, 07:05 AM
HallsofIvy
You don't need to solve for y, use "implicit differentiation".

Since $\displaystyle y^2- 4y- 2x- 1= 0$, $\displaystyle 2yy'- 4y'- 2= 0$. Set y= 1 and solve for y'.

Do it again: $\displaystyle 2y'(y')- 2yy''- 4y''= 2(y')^2- 2yy''- 4y''= 0$. Set y= 1, and y' equal to the value you just found, and solve for y''.