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Y=θ[SIN(INθ)+COS(INθ)],THEN FIND dy÷dθ.
Solution) Y=θ[SIN(INθ)+COS(INθ)]applying u.v rulethen dy÷dθ={[ SIN(INθ)+COS(INθ) ] dθ÷dθ }+ {θ[ d÷dθ{SIN(INθ)+COS(INθ) ] } => SIN(INθ)+COS(INθ) + θ{ (COS(INθ)÷ θ) - (SIN(INθ)÷θ) } θ is canceled and sin(ln θ ) is also canceled then u will get => 2COS(INθ)
Find out the absolute extrema for the given function and interval. g (t ) = 2t 3 + 3t 2 -12t + 4 on [-4, 2] Solution : All we actually need to do here is follow the pr
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the 10 miles assigned to the chess club start at the 10 mile point and go to the 20 mile point when the chess club members have cleaned 5/8 of their 10 mile section between which m
-3+4 #Minimum 100 words accepted#
Derive for the filter from z=a and poles at z=b andz=c, where a, b, c are the real constants the corresponding difference equation. For what values of parameters a, b, and c the fi
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y=Θ[sin(lnΘ)+cos(lnΘ)] dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule) dy/dΘ=2cosΘ.
y=Θ[sin(lnΘ)+cos(lnΘ)]
dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule)
dy/dΘ=2cosΘ.
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