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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)
If tanx+secx=sqr rt 3, 0 Ans) sec 2 x=(√3-tanx) 2 1+tan 2 x=3+tan 2 x-2√3tanx 2√3tanx=2 tanx=1/√3 x=30degree
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1/4+1/2+1/2
Consider the function f(x) = 2x 2 + 1. Find the equation of the tangent to the graph of f(x) at x = 2. [NOTE: when calculating f'(2), use first principles.
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The following table given the these scores and sales be nine salesman during last one year in a certain firm: text scores sales (in 000''rupees) 14 31 19
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Solve by factorization X 2 +(a/a+b + a+b/a)x+1 = 0 X 2 +(a/a+b + a+b/a)x+1 => X 2 +(a/a+b x a+b/ax + a/a+b .a+b/a) => X[x+a/a+b] +a+b/a[a+a*a+b]= 0 => X= -a
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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