higher order derivatives...implicit differentiation

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ku1005

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Oct 27, 2006
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hi, the following is really bothering me because i cant figure out at what step I am going wrong...could someone look at my workin and suggest a solution??thanks

find y'' by implicit differentiation:


x^3 + y^3 = 1

so therefore

y' = - x^2/y^2 (i am quite comfortable this is correct...i hope!)

next i did :

y'' = d (- x^2 / y^2) /dx =

- [(y^2)(2x) - (x^2)(2yy ') ] / y^4

- [2xy-2x^2 y'] / y^3

Then at this stage i SUB y' = (-x^2/y^2)

so

-[2xy-2x^2 (-x^2/y^2)] / y^3

-[2xy + 2x^4y^-2] / y^3

- [2xy + 2x^4] / y^5

HOWEVER....the answer is :

-2x / y^5

....can someone plz tell me where I screwed up!!

thanks heaps
[/u]
 
x^3+y^3=1 take derivative
3x^2 dx/dx + 3y^2 dy/dx=0
3[x^2+y^2 dy/dx]=0
but 3 can't =0
x^2+y^2 dy/dx=0
dy/dx=-x^2 /y^2 answer

dy/dx= -2x/y^5 if this is the answer given, then integration should give us the original. by cross multiplying
y^5 dy=-2x dx integrate
y^6/6=-x^2
y^6+6x^2=0

Arthur
 
A little factoring at the end.

\(\displaystyle \L\\x^{3}+y^{3}=1\)

Skipping ahead, we know \(\displaystyle \L\\\frac{dy}{dx}=\frac{-x^{2}}{y^{2}}\)

\(\displaystyle \L\\y''=-\left(\frac{2x}{y^{2}}-\frac{2x^{2}}{y^{3}}\left(\frac{-x^{2}}{y^{2}}\right)\right)\)

\(\displaystyle =\L\\-\left(\frac{2x}{y^{2}}+\frac{2x^{4}}{y^{5}}\right)\)

Multiply the left side by \(\displaystyle y^{3}\)

\(\displaystyle =\L\\-\left(\frac{y^{3}}{y^{3}}\cdot\frac{2x}{y^{2}}+\frac{2x^{4}}{y^{5}}\right)\)

\(\displaystyle =\L\\-\left(\frac{2xy^{3}+2x^{4}}{y^{5}}\right)\)

Factor:

\(\displaystyle =\L\\-\left(\frac{2x(y^{3}+x^{3})}{y^{5}}\right)\)

See what's in the parentheses?. We know \(\displaystyle x^{3}+y^{3}=1\)

So, you have:

\(\displaystyle =\L\\\frac{-2x}{y^{5}}\)
 
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