Alaina's blog, 3 Oct. 2010

This week in Calculus, we learned implicit derivatives. Basically it is just taking the derivative, but instead of having only one variable, “X”, there are two variables, “X” and “Y”.
Implicit Derivatives are represented as dy/dx with respect to X, or the bottom variable.
To take an Implicit Derivative:
You treat it just like a regular derivative, taking the derivative of each element. When you take the derivative of “Y”, you put dy/dx behind it. Then, you solve for dy/dx. Normally, you end up with a fraction.

Ex 1: X(squared) + Y(squared)=25
2X+2Ydy/dx=0
2Ydy/dx=-2X
dy/dx=-2X/2Y

Ex 2: 2X(squared)+Y(squared)=36X
4X+2Ydy/dx=36
2Ydy/dx=36-4X
dy/dx=(36-4X)/2Y
dy/dx= (18-2X) /Y

Also, we learned Implicit Second Derivatives. Implicit Second Derivatives are represented as
d(squared)y/dx(squared). Basically, you start with a function and follow the rules for regular Implicit Derivatives, which is your first derivative. From there, you take the derivative of your first derivative, second derivative, and plug the first derivative in for dy/dx and simplify.

Ex 1: X(squared) +2Y(squared) =4
2X + 4Ydy/dx =0
4Ydy/dx = -2X
dy/dx = -2X /4Y
dy/dx = -X /2Y
Implicit Second Derivative=> ((2Y)(-1)-[(-X)(2dy/dx)])/(2Y)(squared)
( 2Y+2Xdy/dx)/(2Y)(squared)
(-2Y+2X(-X/2Y))/(2Y)(squared)
(-2Y-(2X(squared)/2Y))/(2Y)(squared)
((-4Y(squared)- 2X(squared) )/2Y)/(2Y)(squared)
(-4Y(squared)-2X(squared )/(4Y)(cubed)
(2( -2Y(squared)-X(squared)))/(2(2Y)(cubed))
(-2Y(squared)-X(squared))/((2Y)(cubed))
X(squared)/Y(cubed)