find d^2y/dx^2, where sqry + 4xy=4?
I believe i can find y' but i can't find y".
I think y' = -(8y^3/2)/(8xsqry+1) but am lost to when it comes to y". I could also be wrong for y', I am just looking for some help !
Cheers
Assuming the initial equation is
\(\displaystyle \sqrt{y} + 4xy = 4\)
then the derivative of y is as you have presented
\(\displaystyle y' = -\dfrac{8 y^{\frac{3}{2}}}{8 x \sqrt{y} + 1} \)
To find the second derivative y'', you can proceed in several ways two of which are:
(1)You could use the original equation to substitute for the value of y
3/2 and \(\displaystyle \sqrt{y}\) and then take derivatives. It would get messy I suspect.
(2)A somewhat simpler way is to write the derivative equation as
\(\displaystyle y' (8 x \sqrt{y} + 1) + 8 y^{\frac{3}{2}} = 0\)
and apply the chain rule as you did before just leaving y' as itself when it occurs. For example if you had
y' x + 2 x
2 y
3/2 = 0
then
y'' x + y' + 4 x y
3/2 + 3 x
2 y
1/2 y'= 0
or
y'' = -[y' (1+ 3 x
2 y
1/2) + 4 x y
3/2 +] / x.
You could then, if necessary, use previous equations to substitute for y and/or y'.
