Holiday Blog #2

Okay for this blog I'm going to review implicit differentiation-and all that means is to take the derivative of an equation with respect to other variables (like y) besides x. So that would make it dy/dx, dt/dx, etc...
So here are some examples:

Ex 1.) Find dy/dx......x^2+y^2=36
*For these you're going to take the derivative all the way across the equation so for the derivative of 36 you'd get zero
*So once you take the derivative you should get:
2x+2ydy/dx=0
*Now you have to move the dy/dx to the other side of the equal sign so you can solve for it
*So now you have:
2ydy/dx = -2x
*Now divide by 2y and should should get this:
dy/dx = -2x/2y which simplifies to -x/y

Ex. 2) Find dy/dx.....x^8+9x+2xy-y^5=4
*First take the derivative of each term all the way across the equation and you should get:
8x^7+9+2(x)(dy/dx)+(y)(1)-5y^4dy/dx = 0
*Now simplify that and you should have:
8x^7+9+2xdy/dx+2y-5y^4dy/dx = 0
*Now take the dy/dx's and move them to one side of the equal sign and the other terms to the other side
*So you should now have:
2xdy/dx-5ydy/dx = -8x^7-2y-9
*Now divide -8x^7-2y-9 by 2x-5y^4
*And your answer is (-8x^7-2y-9)/(2x-5y^4)