# implicit differentiation

• Mar 13th 2010, 09:04 PM
demode
implicit differentiation
Can anyone show me how to find $\displaystyle \frac{dy}{dx}$ for $\displaystyle (x^2+y^2-1)(y-x^2-1) = 0$ at the points (0,1) and (0,0).

I know that I need to use the process of implicit differentiation but I don't know exactly how to apply it here. Do I need to expand the brackets first and then differentiate everything?
• Mar 13th 2010, 09:14 PM
Jhevon
Quote:

Originally Posted by demode
Can anyone show me how to find $\displaystyle \frac{dy}{dx}$ for $\displaystyle (x^2+y^2-1)(y-x^2-1) = 0$ at the points (0,1) and (0,0).

I know that I need to use the process of implicit differentiation but I don't know exactly how to apply it here. Do I need to expand the brackets first and then differentiate everything?

you don't have to expand the brackets. just use the product rule. remember, when differentiating a y-term, you need to attach dy/dx to it.
• Mar 13th 2010, 09:23 PM
Prove It
JHevon brings up a good point, but if you DO expand the brackets, several terms cancel, thus making it easier to calculate the derivative.

$\displaystyle (x^2 + y^2 - 1)(y - x^2 - 1) = 0$

$\displaystyle x^2(y - x^2- 1) + y^2(y - x^2 - 1) - 1(y - x^2 - 1) = 0$

$\displaystyle x^2y - x^4 - x^2 + y^3 - x^2y - y^2 - y + x^2 + 1 = 0$

$\displaystyle -x^4 + y^3 - y^2 - y + 1 = 0$

$\displaystyle y^3 - y^2 - y + 1 = x^4$

$\displaystyle \frac{d}{dx}(y^3 - y^2 - y + 1) = \frac{d}{dx}(x^4)$

$\displaystyle \frac{d}{dy}(y^3 - y^2 - y + 1)\frac{dy}{dx} = 4x^3$

$\displaystyle (3y^2 - 2y - 1)\frac{dy}{dx} = 4x^3$

$\displaystyle (3y + 1)(y - 1)\frac{dy}{dx} = 4x^3$

$\displaystyle \frac{dy}{dx} = \frac{4x^3}{(3y + 1)(y - 1)}$.