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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)
Proof for Properties of Dot Product Proof of u → • (v → + w → ) = u → • v → + u → • w → We'll begin with the three vectors, u → = (u 1 , u 2 , ...
Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C. Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0
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Tests for relative minimum For a relative minimum point there are two tests: i.The first derivative, which is (dy)/(dx) = f´(x) = 0 ii.The second derivative, which i
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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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