Here we go… finally into unknown waters. Wednesday was the first class day where I wasn’t going “Oh YEAH!” and instead was going “Oh… Ok, this is new.” Luckily, I followed the lesson well, and have it down (homework problems helped solidify it for me).

So derivatives combine the stuff we learned about limits (I knew this was going to tie in somehow) and the average rate of change of a function.

Things I found helpful from the lesson:

  • The limit of the average rate of change of a function as h approaches 0 is the instantaneous rate of change.

This can be said in equation form like so: \lim_{h\to0} \dfrac{f(x+h) - f(x)}{h} = f ' (x) where f ' (x) is notation for instantaneous rate of change.

  • The limit of the difference quotient of a function as ha approaches 0 is the derivative of that function.
  • Instantaneous Rate of Change = Derivative
  • The derivative of a function is another function.

Different Notations that all pretty much mean derivative/instantaneous rate of change:

f ' (x)  which is read “f prime of x”

\dfrac{dy}{dx} which is read “dy dx”  (Remember here change in y over change in x? aka slope? yeah… same thing)

D_x read “D sub x”

y' read “y prime”

\dfrac{d}{dx}(f(x)) read “ddx of f(x)”

So basically the average rate of change, slope of the secant line, and the difference quotient are all the same thing.

And the instantaneous rate of change is the same thing as the derivative.

Now that we got all of that out of the way, how the hell do we solve for a derivative? I’ll show you…

But 1st… a quick photo break…. I relate to this so much.

homework-class-test-amidoinitright-oh-god-why-moment

Anyway… I loved this because it spoke to my internal fear of not knowing what’s going on in class. On to me explaining how to find a derivative.

Let’s say you need to find the derivative for a function such as… f(x) = x^2 + 12

First you’re going to want to find the average rate of change for this equation. In this case finding the average rate should go something like this:

1. f(x+h) = (x+h)^2 +12

2. f(x+h) = x^2+2xh+h^2+12

3. f(x+h)-f(x)=x^2+2xh+h^2+12-(x^2 + 12)

4. f(x+h)-f(x)=2xh+h^2

5. \dfrac{f(x+h)-f(x)}{h} = \dfrac{h(2x+h)}{h} which = 2x+h

Now that you have you Difference Quotient you may plug in the awesome new equation for instantaneous rate of change and solve like so:

1.  \lim_{h\to0} \dfrac{f(x+h) - f(x)}{h} = f ' (x)

2. \lim_{h\to0} 2x+h = 2x + 0 = 2x aka f ' (x) = 2x

TADA!!!!!

If any of you don’t understand this please comment and I’ll try to help you the best I can.