Implicit differentiation.

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April 26th 2006, 02:13 PM
dadon

Implicit differentiation.

Hi All!

Having a problem when differentiating for the second time $\frac{d^2y}{dx^2}$ an implicit equation.

I'll give two example problems:

$2x^2 + 2y^2 - 4x + 6 = 0}$
$= 4x + 4y.\frac{dy}{dx} -4 = 0$
$\frac{dy}{dx} = \frac{4 - 4x}{4y}$

Not to sure how to take it from here to do the second derivative $(\frac{d^2y}{dx^2})$

Second problem:

$y^2 + x^2 = 1$
$2y.\frac{dy}{dx} + 2x = 0$
$\frac{dy}{dx} = -\frac{2x}{2y}$

Also not sure how to take it from here to get $\frac{d^2y}{dx^2}$ .

Your help will be appreciated. Thanks

Regards,

dadon
April 26th 2006, 02:18 PM
ThePerfectHacker

Quote:

Originally Posted by dadon

Second problem:

$y^2 + x^2 = 1$
$2y.\frac{dy}{dx} + 2x = 0$
$\frac{dy}{dx} = -\frac{2x}{2y}$

Also not sure how to take it from here to get $\frac{d^2y}{dx^2}$ .

Your help will be appreciated. Thanks

Regards,

dadon

You have,
$x^2+y^2=1$
Thus,
$2x+2yy'=0$ thus, $y'=-\frac{x}{y}$
Thus,
$2+2y''+2y'=0$
Substitute, and divide by 2,
$1+y''-\frac{x}{y}=0$
Thus,
$y''=\frac{x-y}{y}$
April 26th 2006, 02:27 PM
dadon

Sorry I don't get how you get this line:

2 + 2y'' + 2y' = 0

Cheers for the quick response!
April 26th 2006, 02:27 PM
TD!

@ ThePerfectHacker: are you sure about that? Where did your 3rd line come from?

$ \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}\left( { - \frac{x}{y}} \right) = - \frac{{y - xy'}}{{y^2 }} $

Now substituting the expression we have for y' gives

$ \frac{{d^2 y}}{{dx^2 }} = \frac{{ - x\frac{x}{y} - y}}{{y^2 }} = - \frac{{x^2 + y^2 }}{{y^3 }} $

First one goes the same.

Edit: I think you were trying to take the derivative again, after 2x+2yy'=0 but don't forget that y and y' are a function of x. The derivative would give:

$ 2 + 2y'^2 + 2yy'' = 0 \Leftrightarrow y'' = \frac{{ - 1 - y'^2 }}{y} $

Now substituting for y' again yields the same result as above.
April 26th 2006, 03:16 PM
ThePerfectHacker

Sorry, my fault.
My third line should have been, (implicity again),
$2+2y'y'+2yy''=0$
April 27th 2006, 08:26 AM
dadon

Thanks guys!

I will do the second example and get back to you.

Thanks agian.
April 27th 2006, 09:45 AM
TD!

No problem - I'll be glad to check it for you :)
April 27th 2006, 01:32 PM
dadon

re:

Hey cheers!

$ 2x^2 + 2y^2 - 4x + 6 = 0} $
$ 4x + 4y.\frac{dy}{dx} -4 = 0 $
$ \frac{dy}{dx} = \frac{4 - 4x}{4y} $
$\frac{dy}{dx} = -\frac{x}{y}$

This should give the same answer as before as the dy/dx are the same.

This is what I got:

$ \frac{d^2y}{dx^2} $ = $\frac{-y^2 + x^2}{y^3}$

Is the y meant to be negative?
April 27th 2006, 01:35 PM
TD!

Quote:

Originally Posted by dadon

$ \frac{dy}{dx} = \frac{4 - 4x}{4y} $
$\frac{dy}{dx} = -\frac{x}{y}$

This should give the same answer as before as the dy/dx are the same.

What happens here? I think you get

$ \frac{{dy}}{{dx}} = \frac{{4 - 4x}}{{4y}} = \frac{{1 - x}}{y} $

Now take the d/dx again, using the quotient rule and remembering that y = y(x).
April 27th 2006, 01:40 PM
dadon

re:

oh sorry! silly mistake il try agian!
April 27th 2006, 01:50 PM
dadon

so...

$ \frac{d^2y}{dx^2} $ = $\frac{-y^2 - (1-x)^2}{y^3}$

Is the y meant to be negative or as it is squared just change it to positive?
April 27th 2006, 10:30 PM
TD!

Looks good. First we take the derivative again:

$ \left( {\frac{{1 - x}}{y}} \right)^\prime = \frac{{\left( {1 - x} \right)^\prime y - \left( {1 - x} \right)y'}}{{y^2 }} = \frac{{ - y - \left( {1 - x} \right)y'}}{{y^2 }} $

Then substitute for y':

$ \frac{{ - y - \left( {1 - x} \right)y'}}{{y^2 }} = \frac{{ - y - \left( {1 - x} \right)\frac{{1 - x}}{y}}}{{y^2 }} = \frac{{ - y^2 - \left( {1 - x} \right)^2 }}{{y^3 }} $
April 28th 2006, 06:10 AM
dadon

re:

Thanks! Worked out nicely! :)

Kind Regards,

dadon
April 28th 2006, 06:14 AM
TD!

You're welcome :)