How do you find the derivative of #f(x)= (x+sinx)/(cosx) #?

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Answer 1 · 2016-11-03T18:55:35

#f'(x)=(cosx+xsinx+1)/cos^2x#

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

# f(x)=(x+sinx)/cosx#

#u=x+sinx=>u'=1+cosx#

#v=cosx=>v'=-sinx#

#:.f'(x)=(cosx(1+cosx)-(x+sinx)(-sinx))/cos^2x#

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

#f'(x)=(cosx+xsinx+1)/cos^2x#