1. ## Implicit differentiation.

Hi All!

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

I'll give two example problems:

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

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

Second problem:

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

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

Your help will be appreciated. Thanks

Regards,

Second problem:

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

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

Your help will be appreciated. Thanks

Regards,

You have,
$\displaystyle x^2+y^2=1$
Thus,
$\displaystyle 2x+2yy'=0$ thus, $\displaystyle y'=-\frac{x}{y}$
Thus,
$\displaystyle 2+2y''+2y'=0$
Substitute, and divide by 2,
$\displaystyle 1+y''-\frac{x}{y}=0$
Thus,
$\displaystyle y''=\frac{x-y}{y}$

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

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

Cheers for the quick response!

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

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

Now substituting the expression we have for y' gives

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

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:

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

Now substituting for y' again yields the same result as above.

5. Sorry, my fault.
My third line should have been, (implicity again),
$\displaystyle 2+2y'y'+2yy''=0$

6. Thanks guys!

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

Thanks agian.

7. No problem - I'll be glad to check it for you

8. ## re:

Hey cheers!

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

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

This is what I got:

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

Is the y meant to be negative?

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

This should give the same answer as before as the dy/dx are the same.
What happens here? I think you get

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

Now take the d/dx again, using the quotient rule and remembering that y = y(x).

10. ## re:

oh sorry! silly mistake il try agian!

11. so...

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

Is the y meant to be negative or as it is squared just change it to positive?

12. Looks good. First we take the derivative again:

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

Then substitute for y':

$\displaystyle \frac{{ - y - \left( {1 - x} \right)y'}}{{y^2 }} = \frac{{ - y - \left( {1 - x} \right)\frac{{1 - x}}{y}}}{{y^2 }} = \frac{{ - y^2 - \left( {1 - x} \right)^2 }}{{y^3 }}$

13. ## re:

Thanks! Worked out nicely!

Kind Regards,