Question

okay so the derivative of f(x)=cscx is f'(x)=-csc x cot x. Now can someone help me find the x-values where the first derivative is zero or undefined

Answer 1

you have -csc(x)cot(x) = -1/sin(x) * cos(x)/sin(x)
now -1/sin(x) is never equal to 0, so the only way the whole thing will be zero is
if cos(x)/sin(x) = 0
this will only happen when cos(x) = 0
this happens when x = pi/2 + pi* k

Answer 2

that should say \(x=\frac{\pi}{2}\pm\pi*k\) where \(k\in\mathbb{Z}\)

Answer 3

thanks zzr0ck3r but what about undefined?

Answer 4

y' = -cos x/sin^2 x
y' = 0 when cos x = 0-> x = Pi/2 and 3Pi/2 (0 < Pi <2Pi)
Y' undefined when sin x = 0-> x = pi , or 2Pi

Answer 5

its undefined when -1/sin(x) is undefined and when cot(x) = cos(x)/sin(x) is undefined
this happens when sin(x) = 0
and that is when \(x=2\pi*k\) where \(k\in \mathbb{Z}\)

Answer 6

sorry

* and that is when \(x=\pi*k\) where \(k\in \mathbb{Z}\)

Answer 7

@thu1935 its undefined at infinite points

Answer 8

\(\bf -\cfrac{1}{sin(x)=0}\cdot \cfrac{cos(x)}{sin(x)=0}\implies -\cfrac{1}{0}\cdot \cfrac{cos(x)}{0}\)

Answer 9

holy abuse in notation batman

Answer 10

heheh

Answer 11

:P

Answer 12

http://www.wolframalpha.com/input/?i=batman+curve

Answer 13

wow crazy