10/2/10

This week we learned implicit derivatives, derivatives with two variables (x and y) and derivatives "with respect to" a variable, such as dy/dx >> means to take the derivative with respect to y. That goes for any variable (where the top letter is your variable). So keeping that in mind, if you have an equation like x^2+y^2=0, you take the derivative all the way across, but for the y^2 you get 2y dy/dx..because you're taking the derivative with respect to y. This goes for anytime you see a y in an eqation. And remember when you take an implicit derivative, you are solving for dy/dx!!!!!!!!!!..It's pretty easy, except there's a lot of algebra involved, so I think I could used more practice with that. But anyway, here are some examples of what we did this week:

Ex. 1) x^3y^3 - y = x
*The first thing you should notice is that this will be an implicit derivative (duh) because there is both x's and y's in the equation
*The next thing you should notice is that there's a product rule also. x^3y^3..so you should take care of that first
*So you get:
(x^3)(3y^2dy/dx)+(y^3)(3x^2)-dy/dx = 1
*Now you can simplify what you just got and you get
3x^3y^2dy/dx + 3x^2y^3-dy/dx = 1
*Now you have to move the dy/dx's to one side of the equal sign and everything else to the other side
*So you end up with
3x^3y^2dy/dx-dydx = 1-3x^2y^3
*Now you have to solve for dy/dx, so first you would factor out a dy/dx then, you would divide by 3x^3y^2-1
*So your final answer is:
dy/dx = 1-3x^2y^3/3x^3y^2-1

Ex. 2) x^(2/3) + y^(2/3) = 5 (8,1)
Find dy/dx by implicit differentiation and evaluate the derivative at the given point
*First, you start off by taking the derivative like any other problem..putting dy/dx after you take the derivative of the y^2/3..So you have:
2/3(x)^(-1/3) + 2/3(y)^(-1/3)dy/dx = 0
*Now you can simplify and move the 2/3x^(-1/3) to the other side at the same time
*So you have 2/3y^(1/3)dydx = -2/3x^(1/3)
*Finally you divide by 2/3y^(1/3) and you would have to sandwich it..
So after you sandwich it you get -6y^(1/3)/6x^(1/3) and you can cancel out the 6's to make your life easier.
*So you have -y^(1/3)/x^(1/3)
*Unfortunately, you're not done yet. You still have to plug in the point they gave you.
*So you get
-(1)^(1/3)/(8)^(1/3)
= -1/2

Ex. 3) Find an equation of the tangent line of the graph at the given point
(y-3)^2 = 4(x-5) (6,1)
*Firstttttttttt, take the derivative all the way across. **Notice there's a chain rule!
So you get:
2(y-3)(dy/dx) = 4(1) **Remember, leave constants, like 4, out in front
*Now to solve for dy/dx, all you have to do is divide by 2(y-3)
*So you have (4)/2(y-3)...and you can factor out a 2 from the numerator and denominator
*So you get that dy/dx = 2/y-3
*Now you just plug in the point they gave you and you get:
2/-2
=-1 and that's your slope for the equation you're about to set up
*using the point slope formula, like always, you end up with:
y-1=-1(x-6)

**As for what I didn't understand...I'm going to go out on a limb here and say I am LOST with those word problems! (well not totally, like I understand them when Ms Robinson explains them, but I'm not sure I'll be able to work through one by myself just yet) And I could use some more practice with taking the 2nd derivative of implicit derivatives..my algebra is killing me..Other than that, I'd say I'm doing pretty good