Question

can the derivative of csc(x) be equal to
-cot(x)*csc(x) or is it always -csc(x)*cot(x). In my book it says the second one just wanted to know if it makes any difference to use the first one.

Answer 1

i would stick to what your book says is right

Answer 2

The 2nd one. I cant even manipulate the trig proof to get cscx

Answer 3

Yeah but I might blank out on the exam just want to know if it would make a difference because they look the same

Answer 4

I used the quotient rule to get the first one just in case I forget what the book said

Answer 5

No you can only use the the derivative -csc()xcot(x)

(d-dx) csc(x) = (d-dx) 1/sin(x) = ( sin(x) (d-dx) (1) - 1 (d-dx) sin(x) ) / sin2(x) = -cos(x) / sin2(x) = -csc(x)cot(x)

Thats all you need. if you use anything else but -cscxcotx you will get the incorrect answer

Answer 6

thanks lets hope I don't forget