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 # y= 1/sinx (=cscx)#

Method 1 - Chain Rule

Rearrange as # y=(sinx)^-1# and use the chain rule:
# { ("Let "u=sinx, => , (du)/dx=cosx), ("Then "y=u^-1, =>, dy/(du)=-u^-2=-1/u^2 ) :}#

# 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

# { ("Let "u=1, => , (du)/dx=0), ("And "v=sinx, =>, (dv)/dx=cosx ) :}#

# 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 #