Question

Is this right?

Find d^2y/dx^2 in terms of x and y for y^2 = 4x:

y^2 = 4x
2yy' = 4
yy' = 2

y(y'') + y'(y') = 0


sqrt(4x)y'' + (2/y)^2 = 0
y'' = -(2/y)^2 / sqrt(4x)

Answer 1

Oh, I love implicit differentiation, but I've never dealt with the second derivative of it ._.

Answer 2

But wouldn't it start like this?:
\[\LARGE y^2=4x\]
\[\LARGE 2yy'=4\]
\[\LARGE y'=\frac{2}{y}\]

Sorry, I'm rusty on this stuff ._.

Answer 3

2(y^-1)
\[\large y^{-1} + 2y^{-2}y'\] i think

Answer 4

\[\large y'(2y^{-2}) = y^{-1}\]

Answer 5

-y^-1*

Answer 6

Then the simplification.. if im doin it rite

Answer 7

This is messy but:
\[\LARGE y''= \frac{y(-2)'-y'(2)}{y^2}\]
Simplify:
\[\LARGE y''=\frac{0-2y'}{y^2}\]
\[\LARGE y''=-\frac{2y'}{y^2}\]
Now we plug what we got from y' back in:
\[\Huge y''=-\frac{2({\frac{2}{y}})}{y^2}\]
And we simplify from here ._.

Answer 8

\[\large y''(\frac{ 1 }{2y^2 }) = \frac{ -1 }{ y } \]
\[\large y'' = \frac{ -1 }{ y } \times 2y^2\]
\[\large y'' = -2y\]
i think i did something wrong O_O

Answer 9

We learn through mistakes :3