Question

What is the derivative of `(1/sinx)^2`?

Answer 1

`-2cotxcsc^2x`

Explanation:

Note that we can write this more compactly as `csc^2x` as cosec `x`is `1/sinx`.

But to take the derivative this form in terms of sin is more useful. Use the quotient rule for the overall fraction and the chain rule to differentiate `sin^2x`.

`d/dx[1/sin^2x]=(sin^2x*0-1*2sinxcosx)/sin^4x`

`=-(2sinxcosx)/sin^4x=-(2cosx)/sin^3x`

`=-2cotxcsc^2x`