reflection 3

Alright, my third reflection is on Implicit Derivatives.
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)

example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not 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)

and that's it for that problem, it's done.

Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)

First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))

then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)

then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)

after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)

that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.