please show me how to do this!
6 Answers

d/dx (csc x) = csc x cot x
You know that csc(x) = 1/sin(x), so try to make the left side of the equation look like the right side.
d/dx (1/sin(x)) = csc x cot x
So from here just focused on manipulating left side.
Use the quotient rule to find the derivative of 1/sin(x):
[0  1*cos x] / [(sin x)^2] = cos x / (sin x)^2
Split up the (sin x)^2 term into sin x * sin x and rearrange fraction:
[(cos x)/(sin x)] * (1/sin x)
You know that cos x / sin x = cot x & 1/sin x = csc x
Now you have that csc x cot x = csc x cot x.
Hope this helps!

OK check it:
csc (x) = 1/ sin(x)
d/dx 1/sin(x)= [sin(x)(0)  cos(x)(1)]/[sin(x)^2]=
cos(x)/(sin(x)^2)=
cos(x)/sin(x) * 1/sin(x)=
cot(x)csc(x)= csc(x)cot(x)
therefore:
d/dx csc(x)= csc(x)cot(x)

D Dx Csc
Source(s): https://owly.im/a0UDx 
I'm assuming you have knowledge of the power rule and chain rule.
f(x) = csc(x)
f(x) = 1/sin(x)
f(x) = [sin(x)]^(1)
Take the derivative and use the power rule and chain rule.
f'(x) = (1)[sin(x)]^(2) [ cos(x) ]
f'(x) = (cos(x)) / sin^2(x)
Split into two fractions,
f'(x) = [cos(x)/sin(x)] [ 1/sin(x) ]
f'(x) = [ cot(x) ] [ csc(x) ]
f'(x) = csc(x)cot(x)

Derivative of u/v = (u'vuv')/v^2
Derivative of cos^(0.5) x = 0.5 x cos^(0.5)x . ( sin x)
The derivative of root(cos x/sin x) = (0.5cos^(0.5)x . ( sin x) . sin^(0.5)x  cos^(0.5)x . 0.5sin^(0.5)x . cos x) / sin x
=  0.5sin^(0.5)x . cos^(0.5)x (sin^2 x + cos^2 x) / sin x
=  0.5 / (sin^(1.5)x . cos^(0.5)x)
=  0.5 sin^(0.5)x / (sin^2 x cos^(0.5)x)
=  0.5 . csc^2 x . cot^(0.5)x
Nevermind, this is totally wrong.

d/dx(cscx)=d/dx(1/sinx)
quotient rule:
d/dx(1/sinx)=cosx/sin^2x
=(cosx/sinx)(1/sinx)
=(cotx)(cscx)