Question

Q:- Find the derivative of:- X+ Cos x / tan x

Answer 1

x+cosx/tanx
dx/dx + [tanx dcosx/dx-cosx dtanx/dx]/tan²x
1+[tanx(-sinx)-cosx(sec²x)]/tan²x
1+[tanx(-sinx)-secx]/tan²x
1+[-sin²x- cosx(secx)]/cosxtan²x

1+[-sin²x-1]/cosxtan²x
[cosxtan²x-sin²x-1]/cosxtan²x



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Answer 2

Answer:

Let y=x+cosxtanx

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2x

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2dydx=1+tanx(−sinx)−cosxsec2xtan2x

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2dydx=1+tanx(−sinx)−cosxsec2xtan2x=1+−sin2xcosx−cosxsin2x

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2dydx=1+tanx(−sinx)−cosxsec2xtan2x=1+−sin2xcosx−cosxsin2xby multiplying the numerator and denominator by cos2x

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2dydx=1+tanx(−sinx)−cosxsec2xtan2x=1+−sin2xcosx−cosxsin2xby multiplying the numerator and denominator by cos2xdydx=1−cosx1+sin2xsin2x=1−cosx(cosec2x+1)

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2dydx=1+tanx(−sinx)−cosxsec2xtan2x=1+−sin2xcosx−cosxsin2xby multiplying the numerator and denominator by cos2xdydx=1−cosx1+sin2xsin2x=1−cosx(cosec2x+1)dydx=1−cosx−cotxcosecx

Let y=x+cosxtanx ⟹dydx=1+tanxddxcosx−cosxddxtanxtan2xby using ddxuv=vddxu−uddxvv2dydx=1+tanx(−sinx)−cosxsec2xtan2x=1+−sin2xcosx−cosxsin2xby multiplying the numerator and denominator by cos2xdydx=1−cosx1+sin2xsin2x=1−cosx(cosec2x+1)dydx=1−cosx−cotxcosecxHence the required derivative is 1−cosx−cotxcosecx.