Where Memorization Kicks In
In my humble opinion, trigonometry is the hardest subject in algebra. Needless to say. They would eat up a gigantic amount of calculus, and your brain as well! In differential calculus, the problem ain't that serious, but wait till ya get to integrals--
a great place to go for trig review:
Derivatives of 'Elementary' Trigonometric Functions
Here, we try to determine the derivative of the function y=sin(x).
one way of figuring it out would be doing the limit calculations on your own.The other way would be reading the next paragraph.
Countless many have done the calculations before you. the derivative of sin(x) is cos(x). The derivative of cos(x) is -sin(x).
now on your own, you should use the quotient rule and the power rule to obtain the derivatives for respectively, tan(x), sec(x), and csc(x).
After you finish, (Memorize them!)
A hint on tan(x):
Now what is really messy is when the trigonometric function "builds up". That is, what if we encounter some function that is composed of a bunch of sins and cos's mixed up? What if something likes sin(x^2+7)^5 pops up? Things are going to clear up a bit, but still, sometimes, problems tend to remain tough.
This is when your trigonometric skills come in. And memorization follows.
There are really no better ways to calculate cos(2x) than remembering it is 2cos(x)^2-1.(double angle identity) so, when you have the time, eat your way through a trigonometry book.
Derivatives of Inverse Functions
Here, we introduce something that really should be introduced in the chain rule section, but really, it kicks in here. (But knowledge of the chain rule is required, so I recommend giving the above link a click) Soon, we shall introduce the derivatives of inverse trigonometric functions. First, warm yourself up with inverse calculations: e.g. sin(cos-1(x))=sqrt(1-x^2).
Now, think about the derivative of sin-1(x). suppose y=sin-1(x). then we have sin(y)=x. differentiating both sides using the chain rule, we have y'cos(y)=1. or, y'=1/cos(y). Note that we have used y' to denote f'(x). this is a quite common shorthand. If it is a shorthand at all.
The above example can be summarized as f'(x)=1/f'(y). this is true for all inverse functions.
However, in trigonometry, we can further simplify f'(y) by obtaining a relationship between y and x, and substitute y for whatever of x.
For example, up there, we had y'=1/cos(y). Yet, y=sin-1(x). now cos(sin-1(x))=sqrt(1-x^2). Correct me if i am wrong. so now y'=1/sqrt(1-x^2).
Using the same logic, you can deduce some formulas for the rest of the derivatives of the inverse trigonometric functions, with no problem at all!