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fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

d²y/dx²

Find d²y/dx² by implicit ddifferentiation. x^(1/3)+y^(1/3)=4

(1/3)x^(-2/3)+(1/3)y^(-2/3)d/dx=0
d/dx=((-1/3)x^(-2/3))/((1/3)y^(-2/3))
((1/3)y^(2/3))/((-1/3)x^(2/3))
(y^(2/3))/(-x^(2/3))
(-y/x)^(2/3)
d²y/dx² = (2/3)(-y/x)^(-1/3)•((-(d/dx)x+y))/x²

My teacher says the x² in denominator was wrong, or he just didn't get it?
But what do you do after this? I substituted in the y prime and multiplied it out and got:

((-(4/9)x(-y/x)^(-2/3))+(2/3)(-y/x)^(-1/3)y) all over x²

Last edited by fusilli_jerry89 (2006-12-19 13:47:10)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: d²y/dx²

thats how i work it out, although ive only just done the implicit equations chapter so im still not 100% on it.

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: d²y/dx²

luca, you've differentiated them as if they were explicit. That is, you've put a dy/dx on one side and made it all equal to something. That works in explicit differentiation because the equation you're differentiating is always y=..., but it doesn't work here. You've just put the dy/dx on there for no reason.

So your first line of working should read

.

fusilli_jerry, I've followed your workings up until the last line and it's all correct.
I think you may have made a mistake in your workings of d²y/dx² though. It looks like you've treated -y/x as if it was a single variable, which isn't allowed.

You have:
dy/dx = (-y/x)^(2/3)
Split it up: [-y^(2/3)]/[x^(2/3)]

Use the quotient rule: d²y/dx² = [-2/3y^(-1/3)dy/dx*x^(2/3) - 2/3x^(-1/3)*-y^(2/3)]/x^(4/3)

And then to find d²y/dx² in terms of just x and y, you'd substitute in the value of dy/dx that you'd already found.

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Fusilli_Jerry

Guest

Re: d²y/dx²

I redid it and came up with (-2/3)x^(2/3)y^(-1/3)+(2/3)x^(-1/3)y^(2/3) all over x^(4/3)