Question

How do you find the derivative of `1/sinx`?

Answer 1

`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`