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

1 Answer
Dec 5, 2015

-(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.