Blog #6--Stephen

Hope everybody had a good weekend, I know I did. This week was pretty easy, I'm glad it's over, but I am ready for Spirit Week 2010! Anyway, this week in Calculus we learned about implicit derivatives.

When you "take" an implicit derivative, you are "taking" the derivative with two variables. However, the steps do not change when finding the implicit derivative. Work it just like a regular derivative.

d/dx is a derivative with respect to x. You use this when you take the derivative of x. This rule applies to d/dy also. The same concept applies.

When you take a derivative of anything but an x you write dy/dx, dr/dx, ds/dx, the top letter is your variable.

Gather all dy/dx and solve for dy/dx.

Example:

y^3 + y^2 - 5y - x^2 = -4

3y^2(dy/dx) + 2y(dy/dx) - 5(dy/dx) - 2x = 0

3y^2(dy/dx) + 2y(dy/dx) - 5(dy/dx) = 2x

dy/dx(3y^2 + 2y - 5) = 2x

dy/dx = (2x)/(3y^2 + 2y - 5)

dy/dx = (2x)/((3y+5)(y-1))

FINAL ANSWER

It's gonna take a while for me to get used to this complex derivative stuff, but hey, that's what Calc is, COMPLICATED. Oh well, peace out.