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

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Answer 1 · 2015-12-05T16:53:01

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

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

Find each derivative separately.

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

#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#.