Post # 9

Okay, other than taking those exams... this week we learned explicit derivatives!!! Alright, to begin, 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 or y' (I prefer y').
3: Solve for dx/dy or y'
(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(y')+2y(y')-5(y')-2x=0

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

And that's all!!! this problem is finished.

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(y'))=100(y+x(y'))

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

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

after you solved for y', 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)

and I finally learned how to do optimization!!! I got that problem right on the free response part of the exam, but the thing I don't understand, was the second problem on that page. I forgot what it's called, and I don't remember it. All I know is that when I saw it, I didn't know what to do. If anyone remembers try to explain it please. Thanks.