Question

How do you differentiate # g(x) = (4cosx) / (sin^3x) #?

Answer 1

`-(4sin^2x+12cos^2x)/sin^4x`

Explanation:

Use the quotient rule.

`g'(x)=(sin^3xd/dx[4cosx]-4cosxd/dx[sin^3x])/(sin^6x)`

Find each derivative separately.

`d/dx[4cosx]=-4sinx`

Chain rule:

`d/dx[sin^3x]=3sin^2xd/dx[sinx]=3sin^2xcosx`

Plug back in.

`g'(x)=(-4sin^4x-12sin^2xcos^2x)/sin^6x`

`g'(x)=(-sin^2x(4sin^2x+12cos^2x))/sin^6x`

`g'(x)=-(4sin^2x+12cos^2x)/sin^4x`

This can also be written as `-4csc^2x-12cot^2xcsc^2x`.