how does one intuit the derivative of csc x? i can't seem

>>8882830

Like sec but with co- and a minus sign.

d/dx[sec(x)]=sec(x)tan(x)
d/dx[csc(x)]=-csc(x)cot(x)

d/dx[tan(x)]=sec^2(x)
d/dx[cot(x)]=-csc^2(x)

>>8882858
yea but that's just using mnemonic techniques for memorization. sure, a step up from rote memorization. are there not people who just sort of "see" the answer? they imagine what the derivative would look like and realize the shape is -csc(x)cot(x)?

i want to know how those people do that, if that's even something that people new to the subject can do.

>>8882861

It's just the chain rule dude. It's nothing profound.

csc(x) = [sin(x)]^-1
d/dx[csc(x)] = d/dx[[sin(x)]^-1] = - [sin(x)]^-2 * cos(x) = - cos(x)/sin(x) 1/sin(x) = -cot(x)csc(x)

>>8882861
No i assure you almost no one actually would see that the curve is -csc(x)cot(x) , WITHOUT maybe thinking about it and working with trig functions ALOT

>>8882830
So you're not looking for methods to remember it but you're also against having to obtain it by the limit definition (guessing). What exactly are you trying to find?

>>8882870
FFS, when will 4chan fix the lack of latex?
These are painful to read.

>>8882992
[math]Excuse Me?[/math]

>>8882861
No. There's not any useful way to visualize what the graph of a derivative looks like unless it's something incredibly easy like x^2, and even then, without seeing a highly detailed graph you could still guess wrong. Mathematicians aren't in the business of guess work, and I can't imagine any reason why anyone would have the graph of cscxcotx memorized. Derivative graphs aren't all that useful besides a few novel applications.

>>8883060
Aka, stop complaining and memorize your derivative rules. You need to have the simple ones memorized at least for the duration of your calc sequence. After that you can use a calculator or the limit definition

>>8882830
rewrite as 1/(sin(x)) and take the derivative

>>8882992
>>8883055
https://www.mathjax.org/cdn-shutting-down/

>>8882830
Integration by parts