What is the derivative of #-csc(x)#?

1 Answer
Dec 20, 2014

#d/dx (-csc(x)) = -d/dx(csc(x))#
Cosecant is the reciprocal of the sine function, so:
#-d/dx(1/sin(x))#
You should know that the derivative of #1/x = -1/x^2#.
So by the chain rule: #1/f(x) = -1/f(x)^2*d/dxf(x)#

Applying this here:
#-(-1/(sin^2(x))*d/dxsin(x))#
#= -(-1/(sin^2(x))*cos(x))#
# = cos(x)/(sin^2(x))d#
This is already an answer but in most textbooks, you see it in a different form:

# 1/sin(x) * cos(x)/sin(x)#
#= csc(x)*cot(x)#

I hope this helped you in any way.