post # 10

okay, i have like no notes from all week because i was not at school, but i learned some stuff anyway. So at the beginning of the week, the class learned how to take the second implicit derivative.

okay, so as i mentioned in my last post, implicit derivatives have both an x and y variable, and when you take the derivative of y, you note it with dy/dx or with a y'. After you do that, obviously you solve for dy/dx or y'.

alright, so when you take the second derivative, you do the same thing, except when you take the derivative of dy/dx, it becomes d^2y/dx^2

after that you simplify and solve for d^2y/dx^2. and that is all.

ok, first here is an example of a normal implicit derivative: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to note the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0

Then you have to solve for dy'dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)

Now, for the second derivative, you just change the dy/dx into d^2y/dx^2 and take the derivative of the right side of the equation again.

i'm pretty sure that's how it is worked, but then again, i wasn't in class when we learned it.

now, i guess the thing i do not understand anything we learned after the second derivative, and that is really only because i have been absent and have not copied the notes yet and have not tried to learn it.

-mher