9/19/10

Soooo, time for another blog. Starting this week off we learned the difference between average velocity, velocity, and instantaneous velocity. For instance, when you see the words average velocity that means to find the slope. When you see the words instantaneous velocity (or if it's implied) then that means to take the derivative. Then we learned the product and quotient rules used to take derivatives for insanely large functions. I thought that was the easiest thing we learned this week, because all you have to do is remember the formula. On the other hand, some problems get really tricky when it comes down to the simplifying. And also, we learned the derivatives of more trig functions. So here are a few examples of some things we did this week:

Ex 1) f(x) = -1/x [1,2]
^Find the average rate of change of this function over the given interval.
*Alright well first you should notice the key words, average rate of change. This is telling you that you need to used the slope formula
*So the first thing you should do is take your two x-values, 1 & 2, and plug them separately into the f(x) function to get their y-values
*So when you plug in 1 you get: -1/1 = -1
*And when you plug in 2 you get: -1/2 = -1/2
*Then you can plug those numbers into the slope formula y2-y1/x2-x1
*So you get (-1/2 + 1)/2-1 = (1/2)/1
*= 1/2

Ex 2) f(x) = x^3cosx
^Differentiate using the product rule
*Okay so the first thing you should do is decide what two things are being multiplied.....and that would be x^3 and cosx. So you can say x^3 is "f" and cosx is "g"
*Using this formula >> f(x)g'(x) + g(x)f'(x) ..take the derivative
*So this is what you would get when you plug in:
(x^3)(-sinx) + (cosx)(3x^2)
*Then you would distribute that out and get this:
-x^3sinx + 3x^2cosx
*Now you ask yourself if there's anything else you can do to further simplify your answer..
YES!..You can take out a -x^2 to make your life a little bit easier
*And you get:
d/dx = -x^2(xsinx - 3cosx)

Ex 3) f(x) = (x^2-4)/(x-3) c=1
^Differentiate using the quotient rule.
*Sooo first you can label your numerator as "f" and the denominator as "g" to make it easier for you to plug into the formula
*Plugging into this formula >> g(x)f'(x) - f(x)g'(x)/[g(x)]^2 ..you get:
(x-3)(2x) - [(x^2-4)(1)]/(x-3)^2
*Next you can distribute the numbers through. **Don't forget to distribute the negative!
And you get:
2x^2-6x-x^2+4/(x-3)^2
*Now you can combine the like terms in the numerator and you end up with this:
x^2-6x+4/(x-3)^2
*Then you see if there's anything you can do to simplify it further....anddddd you can't so now you can plug in "c" (1) to get your final answer
*So you get: 1^2-6+4/(1-3)^2
*= -1/4

**Okay here's what I didn't quite grasp this week...I could use some further explanation on those velocity word problems, I honestly wouldn't know what to do with them except stare them down on a test paper..and that would probablyyyyyy waste alot of time so I think it would be better if I completely understood how to work them. :)