What is the derivative of #sin^2x/cosx#?

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Answer 1 · 2018-05-28T01:03:03

#f'(x)=((sinx)(2cos^2x+sin^2x))/cos^2x#

#f(x)=(sin^2x)/cosx#

#f'(x)=(cosxtimes2sinxcosx-sin^2xtimes-sinx)/cos^2x#

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

#f'(x)=((sinx)(2cos^2x+sin^2x))/cos^2x#

The quotient rule is given by:

#f(x)=u/v#

#f'(x)=(vu'-uv')/v^2#