Ok, I'm encoding this post in HTML so let me know if there are any problems. I figure I can use some coding to slightly alter the color of the math portions, therefore making them easier to read. Alright, so here we go:

Graph of Y' Stuff

Graphs are pretty important in Calculus as they form an easy allabi when you can't solve a problem. Anyways, there are alot of sweet things to know about a graph for example, if we take the function F(x)=X² who's graph is


then you take the Derivative of the function, which happens to be Y' = 2X¹ and you get this



Now you'll notice several things right away. First, is that the degree of the function is one lower. Second, you'll also notice, and the zero of the derivative is at the minimum of the original function. This can clearly be seen when you graph both of them at the same time



and zoomed in



Example Graphs

I figure I'll use a list instead of a table, the blog site probably won't respond to the coding. The function is in black, whilst the derivative is in red. The original function will come first, then both of then together.

1. F(x)=sinX & Y'=cosX




2. F(x)=X³ + 2X² & Y'=3X² + 4X

So if we can take two rules from all of this:

1. If F(x) is a polynomial, than F'(x) will be one degree lower.


2. If F(x) is continuous and at a maximum or a minimum at A than F'(x) will be 0 at A.
   

Trig Tricks

Alright, even though Trigonometry is a total pain in the ass, we still have to do it. That being said, it's time consuming, stressful, and slightly annoying when you're attempting to find the F'(x) of a trig function. So, there are a couple tricks that make your life a whole lot easier when applied with the other calculus rules. On the left is a function, and on the right in green is the Y' of that function

• F(x)=sinX F'(x)=cosX


• F(x)=cosX F'(x)=-sinX


• F(x)=secX F'(x)=secXtanX


• F(x)=tanX F'(x)=sec²X



• F(x)=cscX F'(x)=-cscXcotX


• F(x)=cotX F'(x)=-csc²X

Below is an example problem of how to use these tricks along with the Quotient Rule:
Your starting function
Y=cosX / (1-sinX)
Apply the Quotient Rule
Y'= (1-sinX)(-sinX) - cosX(-cosX) / (1-sinX)²
Simplify by distributing
Y'=-sinX + sin²X + cos²X / (1 - sinX)²
Next use identies to turn the Sin²X into (1-cos²X) and subtract the cos²X
Y'= -sinX + 1 + - cos²X + cos²X / (1 - sinX)²
Now cancel out a (1 - sinX)
Y'= (1 + sinX) / (1 - sinX)²


Y'= 1 / (1 - sinX)

Alright, that's it

Dalton King