# Implicit differentiation.

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• Apr 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
• Apr 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,
$y''=\frac{x-y}{y}$
• Apr 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!
• Apr 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.
• Apr 26th 2006, 03:16 PM
ThePerfectHacker
Sorry, my fault.
My third line should have been, (implicity again),
$2+2y'y'+2yy''=0$
• Apr 27th 2006, 08:26 AM
dadon
Thanks guys!

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

Thanks agian.
• Apr 27th 2006, 09:45 AM
TD!
No problem - I'll be glad to check it for you :)
• Apr 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?
• Apr 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).
• Apr 27th 2006, 01:40 PM
dadon
re:
oh sorry! silly mistake il try agian!
• Apr 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?
• Apr 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 }}
$
• Apr 28th 2006, 06:10 AM
dadon
re:
Thanks! Worked out nicely! :)

Kind Regards,

dadon
• Apr 28th 2006, 06:14 AM
TD!
You're welcome :)