How do you find the derivative of 1/sinx?

1 Answer
Steve M
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

Related questions
See all questions in Differentiating sin(x) from First Principles
Impact of this question
112853 views around the world
You can reuse this answer
Creative Commons License