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

1 Answer
Jun 22, 2018

-2cotxcsc^2x

Explanation:

Note that we can write this more compactly as csc^2x as cosec xis 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