# Implicit Differentiation

• Oct 25th 2009, 02:30 AM
Danog
Implicit Differentiation
Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

$\frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$

Thanks
• Oct 25th 2009, 02:42 AM
Mush
Quote:

Originally Posted by Danog
Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

$\frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$

Thanks

What is your aim? To find the second derivative of y?
• Oct 25th 2009, 02:51 AM
Danog
Agh, my bad. The questions asks to find the general solution to the equation, and to express y as a function of x.
• Oct 25th 2009, 02:53 AM
ramiee2010
Quote:

Originally Posted by Danog
Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

$\frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$

Thanks

$\frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$
$\frac{dy}{dx} = (3 + 2x)(1 + y^2)$
$\frac{dy}{(1 + y^2)} =(3 + 2x) \cdot dx$
just integrate both side....
• Oct 25th 2009, 02:55 AM
Mush
Quote:

Originally Posted by Danog
Agh, my bad. The questions asks to find the general solution to the equation, and to express y as a function of x.

This isn't really 'implicit differentiation', then. It is the solution of Ordinary Differential Equations!
• Oct 25th 2009, 03:20 AM
Prove It
Quote:

Originally Posted by ramiee2010
$\frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$
$\frac{dy}{dx} = (3 + 2x)(1 + y^2)$
$\frac{dy}{(1 + y^2)} =(3 + 2x) \cdot dx$
just integrate both side....

You use the lazy notation :P.

To be COMPLETELY correct...

$\frac{dy}{dx} = (3 + 2x)(1 + y^2)$

$\frac{1}{1 + y^2}\,\frac{dy}{dx} = 3 + 2x$

$\int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{3 + 2x\,dx}$

$\int{\frac{1}{1 + y^2}\,dy} = 3x + x^2 + C_1$

$\arctan{y} + C_2 = 3x + x^2 + C_1$

$\arctan{y} = 3x + x^2 + C$, where $C = C_1 - C_2$

$y = \tan{(3x + x^2 + C)}$.
• Oct 25th 2009, 03:34 AM
ramiee2010
Quote:

Originally Posted by Prove It
You use the lazy notation :P.

To be COMPLETELY correct...

(Smile) it was just a hint not a complete solution .
• Oct 25th 2009, 03:41 AM
Prove It
Quote:

Originally Posted by ramiee2010
(Smile) it was just a hint not a complete solution .

I'm aware of that. And it was correct. I'm just saying you were using lazy shortcuts that many mathematicians would vomit on...