High Order Deriv.: 2nd deriv. of x^3 + y^3 = 6xy

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Look for the second derivative of:

. . .x^3 + y^3 = 6xy

This is the work I've done:

. . .x^3 + y^3 = 6xy

. . .3x^2 + 3y^2(dy/dx) = 6y + 6x(dy/dx)

. . .3y^2(dy/dx) - 6x(dy/dx) = 6y - 3x^2

. . .dy/dx = (6y - 3x^2) / (3y^2 - 6x)

. . .y' = (2y - x^2) / (y^2 - 2x)

. . .y'' = [(y^2 - 2x)(2(dy/dx) - 2x) - (2y - x^2)(2y(dy/dx) - 2)] / (y^2 - 2x)^2

. . . . .= [(y^2 - 2x)[2(2y - x^2/y^2-2x)-2x] - (2y-x^2)[2y(2y-x^2/y^2-2x)-2]/(y^2-2x)^2

I'm lost on what to do next. I know I have to find a common factor for
everything inside the square brackets, but all the numbers confuse me. Then I'm confused on what to do next. Arggh! This problem is killing me!

I'm sorry if its hard to understand; it's tough transfering it to a computer. Thanks for any help!!
 
I suggest that the differentiations be done first.
\(\displaystyle \L\begin{array}{rcl}
x^3 + y^3 & = & 6xy \\
3x^2 + 3y^2 y' & = & 6y + 6xy' \\
6x + 6y\left( {y'} \right)^2 + 3y^2 y'y'' & = & 6y' + 6y' + 6xy'' \\
6x + 6y\left( {y'} \right)^2 + 3y^2 y'y'' & = & 12y' + 6xy'' \\
\end{array}.\)

Now, solve for y’ is line two.
Then use that to solve for y’’ in the last line.
 
\(\displaystyle \L\begin{array}{rcl}
3x^2 + 3y^2 y' = 6y + 6xy'\quad & \Rightarrow & \quad y' = \frac{{6y - 3x^2 }}{{3y^2 - 6x}} \\
\Rightarrow \quad 6x + 6y\left( {\frac{{6y - 3x^2 }}{{3y^2 - 6x}}} \right)^2 + 3y^2 \left( {\frac{{6y - 3x^2 }}{{3y^2 - 6x}}} \right)y'' & = & 12\left( {\frac{{6y - 3x^2 }}{{3y^2 - 6x}}} \right) + 6xy'' \\
\end{array}\)

Solve for y''.
 
OMG i see it now! Thank you SO much, haha wow i dont know how i didnt understand it the first time you told me. Im going to go work it out. Thanks again!!!!
 
Re: High Order Derivatives

Hello, bandaid-bandet!

\(\displaystyle x^3\,+\,y^3\:=\:6xy\;\;\) Find \(\displaystyle y''\)
Click to expand...

Your work looked fine . . . but you factored illegally.

\(\displaystyle \L x^3\,+\,y^3\:=\:6xy\)

\(\displaystyle \L 3x^2\,+\,3y^2y' \:=\:6y\,+\,6xy'\)

\(\displaystyle \L 3y^2y' \,-\, 6xy' \:= \:6y\,-\,3x^2\)

\(\displaystyle \L 3(y^2\,-\,2x)y' \:= \:3(2y\,-\,x^2)\)

\(\displaystyle \L y' \:=\: \frac{2y\,-\,x^2}{y^2\,-\,2x}\)


\(\displaystyle \L y''\:= \:\frac{(y^2\,-\,2x)(2y'\,-\,2x)\, -\,(2y\,-\,x^2)(2yy'\,-\,2)}{(y^2\,-\,2x)^2} \:=\:2\,\frac{(y^2\,-\,2x)(y'\,-\,x)\,-\,(2y\,-\,x^2)(yy'\,-\,1)}{(y^2\,-\,2x)^2}\)

\(\displaystyle \L y''\:=\:2\,\frac{y^2y'\,-\,xy^2\,-\,2xy'\,+\,2x^2\,-\,2y^2y'\,+\,2y\,+\,x^2yy'\,-\,x^2}{(y^2\,-\,2x)^2}\)

\(\displaystyle \L y''\:=\:2\,\frac{(x^2\,+\,2y\,-\,xy^2)\,+\,(x^2y\,-\,2x\,-\,y^2)y'}{(y^2\,-\,2x)^2}\)

\(\displaystyle \L y''\:=\:2\,\frac{(x^2\,+\,2y\,-\,xy^2)\,+\,(x^2y\,-\,2x\,-\,y^2)\cdot\left(\frac{2y\,-\,x^2}{y^2\,-\,2x}\right)}{(y^2\,-\,2x)^2}\)

Multiply top and bottom by \(\displaystyle (y^2\,-\,2x):\)

\(\displaystyle \L y''\:=\:\frac{(y^2\,-2x)(x^2\,+\,2y\,-\,xy^2)\,+\,(2y\,-\,x^2)(x^2y\,-\,2x\,-\,y^2)}{(y^2\,-\,2x)^3}\)

\(\displaystyle \L y''\:=\:2\,\frac{x^2y^2\,+\,2y^3\,-\,xy^4\,-\,2x^3\,-\,4xy\,+\,2x^2y^2\,+\,2x^2y^2\,-\,4xy\,-\,2y^3\,-\,x^4y\,+\,2x^3\,+\,x^2y^2}{(y^2\,-\,2x)^3}\)

\(\displaystyle \L y''\:=\:2\,\frac{4x^2y^2\,-\,2x^4y\,-\,8xy}{(y^2\,-\,2x)^3} \:=\:2\,\frac{2xy(3xy\,-\,x^3\,-\,4)}{(y^2\,-\,2x)^3}\)

Therefore: \(\displaystyle \L\:y'' \:=\:\frac{4xy(3xy\,-\,x^3\,-\,4)}{(y^2\,-\,2x)^3}\)


But some check my work . . . please!
 
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