How do you differentiate #f(x)=(cos^3x)/sinx# using the quotient rule?

1 Answer
Ratnaker Mehta
Jul 21, 2016

#f'(x)=(-cos^2x(3sin^2x+cos^2x))/sin^2x#.

Explanation:

Before finding the reqd. diifn, we find #d/dxcos^3x# by Chain Rule.

#d/dxcos^3x=d/dx(cosx)^3=3cos^2x*d/dxcosx=-3cos^2xsinx#.

Now, #f'(x)=(sinx*d/dxcos^3x-cos^3x*d/dxsinx)/sin^2x#...[Quotient Rule]

#={(sinx)(-3cos^2xsinx)-(cos^3x)(cosx)}/sin^2x#

#={-3cos^2xsin^2x-cos^4x}/sin^2x#

#=(-cos^2x(3sin^2x+cos^2x))/sin^2x#.

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