How do you find the derivative of #1/sinx#?
1 Answer
Nov 10, 2016
# d/dx (1/sinx)= -cotx cscx #
Explanation:
There are several methods to do this:
Let
Method 1 - Chain Rule
Rearrange as
# dy/dx=(dy/(du))((du)/dx) #
# :. dy/dx = (-1/u^2)(cosx) #
# :. dy/dx = -cosx/sin^2x #
# :. dy/dx = -cosx/sinx * 1/sinx #
# :. dy/dx = -cotx cscx #
Method 2 - Quotient Rule
# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
# :. dy/dx = ( (sinx)(0) - (1)(cosx) ) / (sinx)^2#
# :. dy/dx = -cosx / sin^2x #
# :. dy/dx = -cosx/sinx * 1/sinx #
# :. dy/dx = -cotx cscx #