9/5/10

Okay so this week we had our Chapter 1 limits test..prettyyyyy easy if I do say so myself. Then on Thursday we started derivatives, the long way, which seems easy so far so that's good. We learned the key words for derivatives, such as: dy/dx, y', f'(x), [f(x)], Dx[y]. **And a derivative is also defined as the slope of a tangent line. So I'd say it's pretty important to know these things so you'll be sure to know what a problem is asking you to find. Anyway, I thought pretty much everything we went over this week was easy. So here's a few examples of what we did:

Ex 1) Find the slope of the tangent line of the function f(x)=x^2+3x+4
**First off you have to figure out what the problem is asking you to find. "The slope of the tangent line" is a....DERIVATIVE! So all you have to do for this problem is find the derivative of the function given to you.
*So in order to find the derivative, all you do is plug the function into the derivative formula:
f(x+#x)-f(x)/#x *where lim#x>0...Which means all you do is plug (x+#x) into the equation every time you see an "x" ...(**# will be delta) and then subtract the original equation after. Then simplify the equation, and plug in zero as #x after.
*So you get: (x+#x)^2+3(x+#x)+4-(x^2+3x+4)
*Then expanding that you get:
x^2+2x#x+#x^2+3x+3#x+4-x^2-3x-4 / #x
(*Don't forget to distribute the negative to the original equation)
*Then you can cancel out x^2 and -x^2, 3x and -3x, 4 and -4
*So you have 2x#x+#x^2+3#x/#x
*Now you can factor out #x, since that's the only like term. And that'll cancel out the #x on the denominator also
*So after factoring #x out you get: #x(2x+#x+3)/#x
*Then you can cancel out the #x on the numerator and denominator and plug zero into your final equation >> 2x+#x+3
*So your slope of the tangent line is 2x+3

Ex. 2) Find an equation of the tangent line to the graph of f(x) at the given point.
f(x)=x^3 (2,8)
*The key word of this problem is "tangent line" which tells you you're going to have to find the derivative first
*So you get (x+#x)^3
(x+#x)(x^2+2x#x+#x^2)-x^3 /#x
x^3 + 2x#x + x#x^2 + x^2#x + 2x#x^2 + #x^3 - x^3 / #x
*Now you can cancel out x^3 and -x^3
*So now you have 2x^2#x + x#x^2 + x^2#x + 2x#x^2 + #x^3 / #x
*Next you can factor out a #x
#x(2x^2 + x#x + x^2 2x#x + #x^2) / #x
*then the #x on the numerator and denominator cancel out, and you can plug zero into the equation as #x
*So simplifying that equation you get 3x^2
*Now that you've got the derivative, you have to use the point they gave you, (2,8) to get the slope.
*So to get the slope of 2 (2 being the x term), you plug 2 into 3x^2.
3(2)^2 = 12
*Now to get your final equation you plug the slope and the point into the point slope formula:
y-y1=m(x-x1)
**So you're equation is y-8=12(x-2)

*I basically understood everything we learned this week. The only thing I'm unsure of is how to "discuss the behavior of f at (0,0)" ..like example 4 from our notes.